Topic 7 Inverse Functions

7.1 Inverse Functions - Definition and Properties

Exercise 7.1 If f(x)=x5+x3+x,$f\left(x\right)=x^5+x^3+x,$ find f−1(3)$f^{-1}\left(3\right)$ and (f−1)′(3)${\left(f^{-1}\right)}^{\prime }\left(3\right)$.

Solution.

Since f(f−1(x))=x$f\left(f^{-1}\right(x\left)\right)=x$, by change rule, we have (f−1)′(x)=1f′(f−1(x))$$\left(f^{-1}{)}^{\prime }\right(x)=\frac{1}{f^{\prime }\left(f^{-1}\right(x\left)\right)}$$

To find f′(f−1(3))$f^{\prime }\left(f^{-1}\right(3\left)\right)$, we first find f′$f^{\prime }$: f′(x)=(x5+x3+x)′=5x4+3x2+1$$f^{\prime }\left(x\right)=(x^5+x^3+x{)}^{\prime }=5x^4+3x^2+1$$ which is positive over its domain. So f$f$ is a strictly increasing function and hence one-to-one.

By the definition of inverse function, f−1(3)$f^{-1}\left(3\right)$ is the solution to the equation f(x)=3$f\left(x\right)=3$. Note that f(1)=3$f\left(1\right)=3$, so f−1(3)=1$f^{-1}\left(3\right)=1$.

Put the information together, we find that (f−1)′(3)=1f′(f−1(3))=1f′(1)=15+3+1=19.$$\left(f^{-1}{)}^{\prime }\right(3)=\frac{1}{f^{\prime }\left(f^{-1}\right(3\left)\right)}=\frac{1}{f^{\prime }\left(1\right)}=\frac{1}{5+3+1}=\frac{1}{9}.$$

Exercise 7.2 Find a formula for the inverse of the function f(x)=4x−12x+3$f\left(x\right)=\frac{4x-1}{2x+3}$.

Solution.

Let y=f−1(x)$y=f^{-1}\left(x\right)$. By the definition of f−1(x)$f^{-1}\left(x\right)$, we have x=f(f−1(x))=f(y)=4y−12y+3.$$x=f\left(f^{-1}\right(x\left)\right)=f\left(y\right)=\frac{4y-1}{2y+3}.$$

Solve for y$y$ from this equation, we get y=f−1(x)=−3x+12x−4.$$y=f^{-1}\left(x\right)=-\frac{3x+1}{2x-4}.$$

Exercise 7.3 Find a formula for the inverse of the function f(x)=1+√2+3x$f\left(x\right)=1+\sqrt{2+3x}$.

Solution.

To get the inverse function, solve for y$y$ from the following equation x=1+√2+3y,$$x=1+\sqrt{2+3y},$$ we get f−1(x)=y=13(x−1)2−23.$$f^{-1}\left(x\right)=y=\frac{1}{3}(x-1{)}^2-\frac{2}{3}.$$

Exercise 7.4 Find an explicit formula for f−1$f^{-1}$ where f(x)=x4+1,x≥0$f\left(x\right)=x^4+1,x\ge 0$.

Solution.

Solve for y$y$ from the following equation x=y4+1,$$x=y^4+1,$$ we get f−1(x)=y=4√x−1.$$f^{-1}\left(x\right)=y=\sqrt[4]{4x-1}.$$

Remark: Since the domain of f$f$ is x≥0$x\ge 0$, the range of f−1$f^{-1}$ is y≥0$y\ge 0$. That’s why we take y=4√x−1$y=\sqrt[4]{4x-1}$.

Exercise 7.5 Find (f−1)′(a)${\left(f^{-1}\right)}^{\prime }\left(a\right)$, where f(x)=2x3+3x2+7x+4,a=4$f\left(x\right)=2x^3+3x^2+7x+4,a=4$.

Solution.

Solve f(x)=4$f\left(x\right)=4$, we get f−1(4)=x=0$f^{-1}\left(4\right)=x=0$. As the derivative of f$f$ is f′(x)=6x2+3x+7$f^{\prime }\left(x\right)=6x^2+3x+7$, apply the formula (f−1)′(x)=1f′(f−1(x)),$$\left(f^{-1}{)}^{\prime }\right(x)=\frac{1}{f^{\prime }\left(f^{-1}\right(x\left)\right)},$$ we get (f−1)′(4)=1f′(f−1(4))=1f′(0)=17.$$\left(f^{-1}{)}^{\prime }\right(4)=\frac{1}{f^{\prime }\left(f^{-1}\right(4\left)\right)}=\frac{1}{f^{\prime }\left(0\right)}=\frac{1}{7}.$$

Exercise 7.6 Find (f−1)′(a)${\left(f^{-1}\right)}^{\prime }\left(a\right)$, where f(x)=x3+3sinx+2cosx,a=2$f\left(x\right)=x^3+3\sin x+2\cos x,a=2$.

Solution.

Solve f(x)=2$f\left(x\right)=2$, we get f−1(2)=x=0$f^{-1}\left(2\right)=x=0$. As the derivative of f$f$ is f′(x)=3x2+3cosx−2sinx$f^{\prime }\left(x\right)=3x^2+3\cos x-2\sin x$, apply the formula (f−1)′(x)=1f′(f−1(x)),$$\left(f^{-1}{)}^{\prime }\right(x)=\frac{1}{f^{\prime }\left(f^{-1}\right(x\left)\right)},$$ we get (f−1)′(2)=1f′(f−1(2))=1f′(0)=13.$$\left(f^{-1}{)}^{\prime }\right(2)=\frac{1}{f^{\prime }\left(f^{-1}\right(2\left)\right)}=\frac{1}{f^{\prime }\left(0\right)}=\frac{1}{3}.$$

Exercise 7.7 If g$g$ is an increasing function such that g(2)=8$g\left(2\right)=8$ and g′(2)=5,$g^{\prime }\left(2\right)=5,$ calculate (g−1)′(8)${\left(g^{-1}\right)}^{\prime }\left(8\right)$.

Solution.

Since g(2)=8$g\left(2\right)=8$, g−1(8)=2$g^{-1}\left(8\right)=2$. By the formula (f−1)′(x)=1f′(f−1(x)),$$\left(f^{-1}{)}^{\prime }\right(x)=\frac{1}{f^{\prime }\left(f^{-1}\right(x\left)\right)},$$ we get (g−1)′(8)=1g′(g−1(8))=1g′(2)=15.$$\left(g^{-1}{)}^{\prime }\right(8)=\frac{1}{g^{\prime }\left(g^{-1}\right(8\left)\right)}=\frac{1}{g^{\prime }\left(2\right)}=\frac{1}{5}.$$

Exercise 7.8 If f(x)=∫x3√1+t3dt,$f\left(x\right)={\int }_3^x\sqrt{1+t^3}dt,$ find (f−1)′(0)${\left(f^{-1}\right)}^{\prime }\left(0\right)$.

Solution.

By Fundamental Theorem of Calculus, we know that f′(x)=√1+x3≥0$$f^{\prime }\left(x\right)=\sqrt{1+x^3}\ge 0$$ over its domain x≥−1$x\ge -1$. Then f(x)$f\left(x\right)$ is increasing over [−1,∞]$[-1,\infty ]$. By Fundamental Theorem of Calculus, f(3)=∫33√1+t3dt=0$f\left(3\right)={\int }_3^3\sqrt{1+t^3}dt=0$. So f−1(0)=3$f^{-1}\left(0\right)=3$.

Therefore, (f−1)′(0)=1f′(f−1(0))=1f′(3)=1√28=√714.$$\left(f^{-1}{)}^{\prime }\right(0)=\frac{1}{f^{\prime }\left(f^{-1}\right(0\left)\right)}=\frac{1}{f^{\prime }\left(3\right)}=\frac{1}{\sqrt{28}}=\frac{\sqrt{7}}{14}.$$

7.2 The Natural Logarithmic Function

Exercise 7.9 Find the limit limx→∞[ln(2+x)−ln(1+x)].$$\underset{x\to \infty }{lim}\left[\ln \right(2+x)-\ln (1+x\left)\right].$$

Solution.

limx→∞[ln(2+x)−ln(1+x)]=limx→∞ln(2+x1+x)=ln(limx→∞2+x1+x)=ln1=0$$\begin{array}{cc}& \underset{x\to \infty }{lim}\left[\ln \right(2+x)-\ln (1+x\left)\right]\\ =& \underset{x\to \infty }{lim}\ln \left(\frac{2+x}{1+x}\right)\\ =& \ln \left(\underset{x\to \infty }{lim}\frac{2+x}{1+x}\right)\\ =& \ln 1=0\end{array}$$

Remark: In the second last step, we used the trick that 2+x1+x=2x+11x+1.$$\frac{2+x}{1+x}=\frac{\frac{2}{x}+1}{\frac{1}{x}+1}.$$

Exercise 7.10 Differentiate the function f(x)=√xlnx.$$f\left(x\right)=\sqrt{x}\ln x.$$

Solution.

Apply the chain rule, we get f′(x)=(√xlnx)′=lnx2√x+√xx=√x(lnx+2)2x.$$f^{\prime }\left(x\right)=(\sqrt{x}\ln x{)}^{\prime }=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}=\frac{\sqrt{x}(\ln x+2)}{2x}.$$

Exercise 7.11 Differentiate the function f(x)=ln(sin2x).$$f\left(x\right)=\ln \left({\sin }^{2}x\right).$$

Solution.

Apply the chain rule twice, we get f′(x)=1sin2x(sin2x)′=1sin2x(2sinx)(sinx)′=2tanx.$$f^{\prime }\left(x\right)=\frac{1}{{\sin }^{2}x}({\sin }^{2}x{)}^{\prime }=\frac{1}{{\sin }^{2}x}(2\sin x\left)\right(\sin x{)}^{\prime }=2\tan x.$$

Exercise 7.12 Differentiate the function y=1lnx.$$y=\frac{1}{\ln x}.$$

Solution.

Apply the chain rule, we get f′(x)=−1(lnx)2(lnx)′=−1x(lnx)2.$$f^{\prime }\left(x\right)=-\frac{1}{(\ln x{)}^2}(\ln x{)}^{\prime }=-\frac{1}{x(\ln x{)}^2}.$$

Exercise 7.13 Differentiate the function y=(lntanx)2.$$y=(\ln \tan x{)}^2.$$

Solution.

Apply the chain rule twice, we get f′(x)=2(lntanx)(lntanx)′=2(lntanx)tanx(tanx)′=2sec2x(lntanx)tanx.$$f^{\prime }\left(x\right)=2\left(\ln \tan x\right)(\ln \tan x{)}^{\prime }=\frac{2\left(\ln \tan x\right)}{\tan x}(\tan x{)}^{\prime }=\frac{2{\sec }^{2}x\left(\ln \tan x\right)}{\tan x}.$$

Exercise 7.14 Find f′′(e)$f^{\prime \prime }\left(e\right)$, where f(x)=lnxx.$$f\left(x\right)=\frac{\ln x}{x}.$$

Solution.

Apply the quotient rule and simplify, we get f′(x)=1−lnxx2$$f^{\prime }\left(x\right)=\frac{1-\ln x}{x^2}$$

Apply the quotient rule again and simplify, we get f″(x)=2lnx−3x3$$f^{\prime \prime }\left(x\right)=\frac{2\ln x-3}{x^3}$$

Therefore, f″(e)=2lne−3e3=2⋅1−3e3=−1e3.$$f^{\prime \prime }\left(e\right)=\frac{2\ln e-3}{e^3}=\frac{2\cdot 1-3}{e^3}=-\frac{1}{e^3}.$$

Exercise 7.15 Use logarithmic differentiation to find the derivative of the function y=√x−1x4+1.$$y=\sqrt{\frac{x-1}{x^4+1}}.$$

Solution.

Apply ln$\ln$ to both sides, we get lny=ln(√x−1x4+1)=12ln(x−1)−12ln(x4+1).$$\ln y=\ln \left(\sqrt{\frac{x-1}{x^4+1}}\right)=\frac{1}{2}\ln (x-1)-\frac{1}{2}\ln (x^4+1).$$

Differentiate both sides, we get 1y⋅y′=12(x−1)−4x32(x4+1).$$\frac{1}{y}\cdot y^{\prime }=\frac{1}{2(x-1)}-\frac{4x^3}{2(x^4+1)}.$$

Multiply both sides by y=√x−1x4+1$y=\sqrt{\frac{x-1}{x^4+1}}$, we get y′=(12(x−1)−4x32(x4+1))√x−1x4+1$$y^{\prime }=(\frac{1}{2(x-1)}-\frac{4x^3}{2(x^4+1)})\sqrt{\frac{x-1}{x^4+1}}$$

Exercise 7.16 Evaluate the integral ∫423xdx.$${\int }_2^4\frac{3}{x}dx.$$

Solution.

∫423xdx=3∫421xdx=3(ln|x|)|42=3(ln4−ln2)=3ln2.$$\begin{array}{cc}& {\int }_2^4\frac{3}{x}dx\\ =& 3{\int }_2^4\frac{1}{x}dx\\ =& 3\left(\ln |x|\right){|}_2^4\\ =& 3(\ln 4-\ln 2)=3\ln 2.\end{array}$$

Exercise 7.17 Evaluate the integral ∫21dt8−3t.$${\int }_1^2\frac{dt}{8-3t}.$$

Solution.

We find the anti-derivative first: ∫dt8−3tu=8−3t=−13∫duu=−13(ln|u|)+C=−13(ln|8−3t|)+C.$$\begin{array}{cc}& \int \frac{dt}{8-3t}\\ \stackrel{u=8-3t}{=}& -\frac{1}{3}\int \frac{du}{u}\\ =& -\frac{1}{3}\left(\ln |u|\right)+C\\ =& -\frac{1}{3}(\ln |8-3t|)+C.\end{array}$$

By Fundamental Theorem of Calculus, we get ∫21dt8−3t=−13(ln|8−3⋅2|−ln|8−3⋅1|)=ln5−ln23.$${\int }_1^2\frac{dt}{8-3t}=-\frac{1}{3}(\ln |8-3\cdot 2|-\ln |8-3\cdot 1|)=\frac{\ln 5-\ln 2}{3}.$$

Exercise 7.18 Evaluate the integral ∫cosx2+sinxdx.$$\int \frac{\cos x}{2+\sin x}dx.$$

Solution.

Let u=2+sinx$u=2+\sin x$. We find that du=cosxdx$du=\cos xdx$. Then ∫cosx2+sinxdx=∫1udu=ln|u|+C=ln(2+sinx)+C.$$\int \frac{\cos x}{2+\sin x}dx=\int \frac{1}{u}du=\ln |u|+C=\ln (2+\sin x)+C.$$

Remark: In the last step, we remove the absolute value sign because 2+sinx≥1$2+\sin x\ge 1$.

7.3 The Natural Exponential Function

Exercise 7.19 Find the domain of the function f(x)=√3−e2x.$$f\left(x\right)=\sqrt{3-e^{2x}}.$$

Solution.

The domain of the function is given by the inequality 3−e2x≥0,$$3-e^{2x}\ge 0,$$ which is equivalent to e2x≥3.$$e^{2x}\ge 3.$$

As the natural logarithmic function is strict increasing, take ln$\ln$ of both sides, we get 2x≥ln3x≥12ln3.$$\begin{array}{cc}2x\ge & \ln 3\\ x\ge & \frac{1}{2}\ln 3.\end{array}$$

Exercise 7.20 Find the limit limx→∞(e−2xcosx).$$\underset{x\to \infty }{lim}\left(e^{-2x}\cos x\right).$$

Solution.

Since −1≤cosx≤1$-1\le \cos x\le 1$ and e−2x>0$e^{-2x}>0$, we have the following inequality −e−2x≤e−2xcosx≤e−2x.$$-e^{-2x}\le e^{-2x}\cos x\le e^{-2x}.$$

By Squeeze theorem and the fact that limx→∞e−2x=0${lim}_{x\to \infty }{e}^{-2x}=0$, we know that limx→∞(e−2xcosx)=0.$$\underset{x\to \infty }{lim}\left(e^{-2x}\cos x\right)=0.$$

Exercise 7.21 Differentiate the function f(t)=sin(et)+esint.$$f\left(t\right)=\sin \left(e^t\right)+e^{\sin t}.$$

Solution.

Apply chain rule to both terms: f′(x)=(sin(et))′+(esint)′=cos(et)et+esint(cost)=etcos(et)+esintcost.$$\begin{array}{cc}{f}^{\prime }\left(x\right)& ={\left(\sin \right(e^t\left)\right)}^{\prime }+{\left(e^{\sin t}\right)}^{\prime }\\ & =\cos \left(e^t\right)e^t+e^{\sin t}\left(\cos t\right)\\ & =e^t\cos \left(e^t\right)+e^{\sin t}\cos t.\end{array}$$

Exercise 7.22 Differentiate the function y=cos(1−e2x1+e2x).$$y=\cos \left(\frac{1-e^{2x}}{1+e^{2x}}\right).$$

Solution.

Note that 1−e2x1+e2x=21+e2x−1.$$\frac{1-e^{2x}}{1+e^{2x}}=\frac{2}{1+e^{2x}}-1.$$

Apply chain rule three times: y′=−sin(1−e2x1+e2x)⋅(21+e2x−1)′=−sin(1−e2x1+e2x)⋅(−2(1+e2x)2)⋅(1+e2x)′=sin(1−e2x1+e2x)⋅(2(1+e2x)2)⋅2e2x=4e2x(1+e2x)2sin(1−e2x1+e2x).$$\begin{array}{cc}{y}^{\prime }& =-\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\cdot {(\frac{2}{1+e^{2x}}-1)}^{\prime }\\ & =-\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\cdot (-\frac{2}{(1+e^{2x}{)}^2})\cdot {(1+e^{2x})}^{\prime }\\ & =\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\cdot \left(\frac{2}{(1+e^{2x}{)}^2}\right)\cdot 2e^{2x}\\ & =\frac{4e^{2x}}{(1+e^{2x}{)}^2}\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right).\end{array}$$

Exercise 7.23 Find an equation of the tangent line to the curve xey+yex=1$$x{e}^y+y{e}^x=1$$ at the point (0,1)$(0,1)$.

Solution.

The slope of the tangent line is given by y′(0)$y^{\prime }\left(0\right)$. So we first need to find y′(x)$y^{\prime }\left(x\right)$.

Differentiate both side with respect to x$x$, we get ey+xeyy′+y′ex+yex=0.$$e^y+x{e}^y{y}^{\prime }+y^{\prime }{e}^x+y{e}^x=0.$$

Solve for y′$y^{\prime }$, we obtain y′=−ey+yexxey+ex$$y^{\prime }=-\frac{e^y+y{e}^x}{x{e}^y+e^x}$$

Plug (0,1)$(0,1)$ into the right hand side, we get the slope of the tangent line m=y′(0)=−e+11=−(e+1).$$m=y^{\prime }\left(0\right)=-\frac{e+1}{1}=-(e+1).$$

So the tangent line is defined by y=−(e+1)x+1.$$y=-(e+1)x+1.$$

Exercise 7.24 Find the absolute maximum value of the function f(x)=x−ex.$$f\left(x\right)=x-e^x.$$

Solution.

The function f$f$ is differentiable over all real numbers. To find the absolute maximum, we study the monotonicity of the function.

The first derivative is f′(x)=1−ex$f^{\prime }\left(x\right)=1-e^x$. Since ex>1$e^x>1$ for x>0$x>0$ and ex<1$e^x<1$ for x<0$x<0$, we know that f′(x)<0$f^{\prime }\left(x\right)<0$ for x>0$x>0$ and f′(x)>0$f^{\prime }\left(x\right)>0$ for x<0$x<0$.

The information shows that f(x)$f\left(x\right)$ is increasing over (−∞,0)$(-\infty ,0)$ and decreasing over (0,∞)$(0,\infty )$.

Thus, f(x)$f\left(x\right)$ has a maximum f(0)=0−e0=−1$f\left(0\right)=0-e^0=-1$.

Exercise 7.25 Evaluate the integral ∫ex√1+exdx.$$\int e^x\sqrt{1+e^x}dx.$$

Solution.

∫ex√1+exdx=∫√uduu=1+ex=23u32+C=23(1+ex)32+C.$$\begin{array}{cc}& \int e^x\sqrt{1+e^x}dx\\ =& \int \sqrt{u}duu=1+e^x\\ =& \frac{2}{3}{u}^{\frac{3}{2}}+C\\ =& \frac{2}{3}(1+e^x{)}^{\frac{3}{2}}+C.\end{array}$$

Exercise 7.26 Evaluate the integral ∫(ex+e−x)2dx.$$\int {(e^x+e^{-x})}^{2}dx.$$

Solution.

∫(ex+e−x)2dx=∫(e2x+2+e−2x)dx=12e2x+2x−12e−2x+C.$$\begin{array}{cc}& \int {(e^x+e^{-x})}^{2}dx\\ =& \int (e^{2x}+2+e^{-2x})dx\\ =& \frac{1}{2}{e}^{2x}+2x-\frac{1}{2}{e}^{-2x}+C.\end{array}$$

7.4 General Logarithmic and Exponential Functions

Exercise 7.27 Differentiate the function f(x)=x5+5x.$$f\left(x\right)=x^5+5^x.$$

Solution.

f′(x)=5x+5xln5.$$f^{\prime }\left(x\right)=5x+5^x\ln 5.$$

Exercise 7.28 Differentiate the function g(x)=xsin(2x).$$g\left(x\right)=x\sin \left(2^x\right).$$

Solution.

f′(x)=sin(2x)+xcos(2x)(2x)′=sin(2x)+x2xcos(2x)ln2.$$\begin{array}{cc}{f}^{\prime }\left(x\right)=& \sin \left(2^x\right)+x\cos \left(2^x\right)(2^x{)}^{\prime }\\ =& \sin \left(2^x\right)+x{2}^x\cos \left(2^x\right)\ln 2.\end{array}$$

Exercise 7.29 Differentiate the function f(x)=log5(xex).$$f\left(x\right)={\log }_5\left(x{e}^x\right).$$

Solution.

f′(x)=1ln5(lnx+x)′=1ln5(1x+1)=x+1xln5.$$\begin{array}{cc}{f}^{\prime }\left(x\right)=& \frac{1}{\ln 5}(\ln x+x{)}^{\prime }\\ =& \frac{1}{\ln 5}(\frac{1}{x}+1)\\ =& \frac{x+1}{x\ln 5}.\end{array}$$

Exercise 7.30 Differentiate the function y=xcosx.$$y=x^{\cos x}.$$

Solution.

f′(x)=(ecosxlnx)′=xcosx(cosxlnx)′=xcosx(−sinxlnx+cosxx)=(cosx−xsinxlnx)xcosxx.$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =(e^{\cos x\ln x}{)}^{\prime }=x^{\cos x}(\cos x\ln x{)}^{\prime }\\ & =x^{\cos x}(-\sin x\ln x+\frac{\cos x}{x})\\ & =\frac{(\cos x-x\sin x\ln x)x^{\cos x}}{x}.\end{array}$$

Exercise 7.31 Evaluate the integral ∫(x5+5x)dx.$$\int (x^5+5^x)dx.$$

Solution.

∫(x5+5x)dx=x66+5xln5+C.$$\begin{array}{cc}& \int (x^5+5^x)dx\\ =& \frac{x^6}{6}+\frac{5^x}{\ln 5}+C.\end{array}$$

Exercise 7.32 Evaluate the integral ∫log10xxdx.$$\int \frac{{\log }_{10}x}{x}dx.$$

Solution.

∫log10xxdx=1ln10∫lnxxdx=1ln10∫lnx d(lnx)u=lnx=(lnx)22ln10+C.$$\begin{array}{cc}& \int \frac{{\log }_{10}x}{x}dx\\ =& \frac{1}{\ln 10}\int \frac{\ln x}{x}dx\\ =& \frac{1}{\ln 10}\int \ln xd\left(\ln x\right)u=\ln x\\ =& \frac{(\ln x{)}^2}{2\ln 10}+C.\end{array}$$

Exercise 7.33 Evaluate the integral ∫102x2x+1dx.$${\int }_0^1\frac{2^x}{2^x+1}dx.$$

Solution.

Let u=2x+1$u=2^x+1$, then du=2xln2dx$du=2^x\ln 2dx$. Therefore, ∫2x2x+1dx=1ln2∫1udu=lnuln2+C=ln(2x+1)ln2+C.$$\begin{array}{cc}& \int \frac{2^x}{2^x+1}dx\\ =& \frac{1}{\ln 2}\int \frac{1}{u}du\\ =& \frac{\ln u}{\ln 2}+C\\ =& \frac{\ln (2^x+1)}{\ln 2}+C.\end{array}$$

By FTC, ∫102x2x+1dx=ln(21+1)ln2−ln(20+1)ln2=ln3−ln2ln2.$$\begin{array}{cc}{\int }_0^1\frac{2^x}{2^x+1}dx=& \frac{\ln (2^1+1)}{\ln 2}-\frac{\ln (2^0+1)}{\ln 2}\\ =& \frac{\ln 3-\ln 2}{\ln 2}.\end{array}$$

Exercise 7.34 Find the area of the region bounded by the curves y=2x$y=2^x$, y=5x,x=−1,$y=5^x,x=-1,$ and x=1$x=1$

Solution.

Note that 5x>2x$5^x>2^x$ for x>0$x>0$ and 2x>5x$2^x>5^x$ for x<0$x<0$

So the area of the region is given by ∫0−1(2x−5x)dx+∫10(5x−2x)dx=(2xln2−5xln5)|0−1+(5xln5−2xln2)|10=165ln2−12ln5ln2ln5$$\begin{array}{cc}& {\int }_{-1}^0(2^x-5^x)dx+{\int }_0^1(5^x-2^x)dx\\ =& (\frac{2^x}{\ln 2}-\frac{5^x}{\ln 5}){|}_{-1}^0+(\frac{5^x}{\ln 5}-\frac{2^x}{\ln 2}){|}_0^1\\ =& \frac{\frac{16}{5}\ln 2-\frac{1}{2}\ln 5}{\ln 2\ln 5}\end{array}$$

7.5 Exponential Growth and Decay

Exercise 7.35 A sample of tritium-3 decayed to 94.5% of its original amount after a year.

1. What is the half-life of tritium-3?
2. How long would it take the sample to decay to 20% of its original amount?

Solution.

The decay of tritium-3 can be modeled by the function m(t)=m(0)ekt.$$m\left(t\right)=m\left(0\right)e^{kt}.$$

We know that m(1)=0.945m(0)$m\left(1\right)=0.945m\left(0\right)$ which implies that ek=0.945$e^k=0.945$. So k=ln(0.945)$k=\ln \left(0.945\right)$.

The half-life is the solution of the equation 0.5=etln(0.945).$$0.5=e^{t\ln \left(0.945\right)}.$$ Taking ln$\ln$ of both sides and solve for t$t$, we get the half-life is ln(0.5)ln(0.945)≈12.25$\frac{\ln \left(0.5\right)}{\ln \left(0.945\right)}\approx 12.25$ years.

Similarly, it takes the sample ln(0.2)ln(0.945)≈28.45$\frac{\ln \left(0.2\right)}{\ln \left(0.945\right)}\approx 28.45$ years to decay to 20%.

Exercise 7.36 A curve passes through the point (0,5)$(0,5)$ and has the property that the slope of the curve at every point P$P$ is twice the y$y$-coordinate of P$P$. What is the equation of the curve?

Solution.

The slope of the line is given by the derivative y′$y^{\prime }$. From the given information, we know that y′=2y$y^{\prime }=2y$ which has the solution y(x)=Ce2x$y\left(x\right)=C{e}^{2x}$.

Since the point (0,5)$(0,5)$ is on the curve, we have 5=y(0)=Ce0$5=y\left(0\right)=C{e}^0$. So C=5$C=5$ and the equation of the curve is y=5e2x.$$y=5e^{2x}.$$

Exercise 7.37 A freshly brewed cup of tea has temperature 97∘${}^{\circ }$C. The temperature of the room is 22∘${}^{\circ }$C. After 10 minutes, the temperature of the coffee is 47∘${}^{\circ }$C. Find the temperature of the coffee after 20 minutes. By Newton’s cooling law, the temperature of the coffee satisfies the equation y′(t)=k(y(t)−R)$y^{\prime }\left(t\right)=k\left(y\right(t)-R)$, where R$R$ is the room temperature.

Solution.

Since R=22$R=22$ and y(0)=97$y\left(0\right)=97$, by Newton’s cooling law, the temperature of the coffee can be modeled by y(t)=(98−23)ekt+23=75ekt+23$y\left(t\right)=(98-23)e^{kt}+23=75e^{kt}+23$.

We also know that y(10)=47$y\left(10\right)=47$ which gives us an equation 75e10k+22=47.$$75e^{10k}+22=47.$$

Solve for k$k$, we get t=−ln310.$t=-\frac{\ln 3}{10}.$

Therefore, y(20)=75e−2ln3+22=75⋅(eln3)−2+22=913≈30.3∘C.$$\begin{array}{cc}y\left(20\right)=& 75e^{-2\ln 3}+22\\ =& 75\cdot {\left(e^{\ln 3}\right)}^{-2}+22\\ =& \frac{91}{3}\approx {30.3}^{\circ }\text{C}.\end{array}$$

Exercise 7.38 How long will it take an investment to double in value if the interest rate is 6% compounded continuously?

Solution.

The balance function of an investment with an annual rate 6% compounded continuous is given by B(t)=Pe0.06t.$$B\left(t\right)=P{e}^{0.06t}.$$

When the balance is doubled, it satisfies the equation 2P=Pe0.06.$$2P=P{e}^{0.06}.$$

Solve the equation, we get that it takes about t=ln2/0.06≈11.55$t=\ln 2/0.06\approx 11.55$ years that the investment will be doubled.

7.6 Inverse Trigonometric Functions

Exercise 7.39 Find the derivative of the function

y=tan−1(x2)$$y={\tan }^{-1}\left(x^2\right)$$

Solution.

By the chain rule, we get y′=2x1+x4.$$y^{\prime }=\frac{2x}{1+x^4}.$$

Exercise 7.40 Find the derivative of the function

y=sin−1(2x+1)$$y={\sin }^{-1}(2x+1)$$

Solution.

By the chain rule, we get y′=2√1−(2x+1)2.$$y^{\prime }=\frac{2}{\sqrt{1-(2x+1{)}^2}}.$$

Exercise 7.41 Find the derivative of the function

f(θ)=arcsin√sinθ$$f\left(\theta \right)=\arcsin \sqrt{\sin \theta }$$

Solution.

Apply the chain rule twice, we get y′=1√1−sinθ⋅12√sinθ⋅cosθ.$$y^{\prime }=\frac{1}{\sqrt{1-\sin \theta }}\cdot \frac{1}{2\sqrt{\sin \theta }}\cdot \cos \theta .$$

Exercise 7.42 Find the derivative of the function

y=arctan√1−x1+x$$y=\arctan \sqrt{\frac{1-x}{1+x}}$$

Solution.

Note that 1−x1+x=21+x−1$\frac{1-x}{1+x}=\frac{2}{1+x}-1$.

Apply the chain rule three times, we get y′=11+1−x1+x⋅12√1−x1+x⋅(−2(x+1)2)=−12(x+1)√1+x1−x.$$\begin{array}{cc}{y}^{\prime }=& \frac{1}{1+\frac{1-x}{1+x}}\cdot \frac{1}{2\sqrt{\frac{1-x}{1+x}}}\cdot (-\frac{2}{(x+1{)}^2})\\ =& -\frac{1}{2(x+1)}\sqrt{\frac{1+x}{1-x}}.\end{array}$$

Exercise 7.43 Evaluate the integral

∫√31/√381+x2dx$${\int }_{1/\sqrt{3}}^{\sqrt{3}}\frac{8}{1+x^2}dx$$

Solution.

∫√31/√381+x2dx=8arctan(x)|√31/√3=8(π3−π6)=4π3.$$\begin{array}{cc}{\int }_{1/\sqrt{3}}^{\sqrt{3}}\frac{8}{1+x^2}dx=& 8\arctan \left(x\right){|}_{1/\sqrt{3}}^{\sqrt{3}}\\ =& 8(\frac{\pi }{3}-\frac{\pi }{6})\\ =& \frac{4\pi }{3}.\end{array}$$

Exercise 7.44 Evaluate the integral

∫1/20sin−1x√1−x2dx$${\int }_0^{1/2}\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx$$

Solution.

Let u=sin−1x$u={\sin }^{-1}x$. Then ∫sin−1x√1−x2dx=∫udu=12u2+C=12(sin−1x)2+C.$$\begin{array}{c}\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx=\int udu=\frac{1}{2}{u}^2+C=\frac{1}{2}({\sin }^{-1}x{)}^2+C.\end{array}$$

Therefore, ∫1/20sin−1x√1−x2dx=12(sin−1(1/2))2−12(sin−10)2=12(π6)2=π72.$$\begin{array}{cc}& {\int }_0^{1/2}\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx\\ =& \frac{1}{2}\left({\sin }^{-1}\right(1/2){)}^2-\frac{1}{2}({\sin }^{-1}0{)}^2\\ =& \frac{1}{2}{\left(\frac{\pi }{6}\right)}^2=\frac{\pi }{72}.\end{array}$$

Exercise 7.45 Evaluate the integral

∫π/20sinx1+cos2xdx$${\int }_0^{\pi /2}\frac{\sin x}{1+{\cos }^{2}x}dx$$

Solution.

Let u=cosx$u=\cos x$. Then ∫sinx1+cos2xdx=∫11+u2du=arctan(u)+C=arctan(cosx)+C.$$\begin{array}{cc}& \int \frac{\sin x}{1+{\cos }^{2}x}dx\\ =& \int \frac{1}{1+u^2}du\\ =& \arctan \left(u\right)+C=\arctan \left(\cos x\right)+C.\end{array}$$

Therefore, ∫π/20sinx1+cos2xdx=arctan(cos(π/2))−arctan(cos0)=arctan(0)−arctan(1)=−π4.$$\begin{array}{cc}& {\int }_0^{\pi /2}\frac{\sin x}{1+{\cos }^{2}x}dx\\ =& \arctan \left(\cos \right(\pi /2\left)\right)-\arctan \left(\cos 0\right)\\ =& \arctan \left(0\right)-\arctan \left(1\right)=-\frac{\pi }{4}.\end{array}$$

Exercise 7.46 Evaluate the integral

∫e2x√1−e4xdx$$\int \frac{e^{2x}}{\sqrt{1-e^{4x}}}dx$$

Solution.

Let u=e2x$u=e^{2x}$. Then du=2e2xdx$du=2e^{2x}dx$ and ∫e2x√1−e4xdx=12∫1√1−u2du=12arcsin(u)+C=12arcsin(e2x)+C.$$\begin{array}{cc}& \int \frac{e^{2x}}{\sqrt{1-e^{4x}}}dx\\ =& \frac{1}{2}\int \frac{1}{\sqrt{1-u^2}}du\\ =& \frac{1}{2}\arcsin \left(u\right)+C=\frac{1}{2}\arcsin \left(e^{2x}\right)+C.\end{array}$$

Exercise 7.47 Find the limit

limx→∞arccos(1+x21+2x2)$$\underset{x\to \infty }{lim}\arccos \left(\frac{1+x^2}{1+2x^2}\right)$$

Solution.

Because cosx$\cos x$ is continuous over its domain [-1, 1] and 0≤1+x21+2x2≤1$0\le \frac{1+x^2}{1+2x^2}\le 1$.

Then limx→∞arccos(1+x21+2x2)=arccos(limx→∞1+x21+2x2)=arccos(12)=π3.$$\begin{array}{cc}\underset{x\to \infty }{lim}\arccos \left(\frac{1+x^2}{1+2x^2}\right)=& \arccos \left(\underset{x\to \infty }{lim}\frac{1+x^2}{1+2x^2}\right)\\ =& \arccos \left(\frac{1}{2}\right)=\frac{\pi }{3}.\end{array}$$

Exercise 7.48 Find the limit

limx→0+tan−1(lnx)$$\underset{x\to 0^{+}}{lim}{\tan }^{-1}\left(\ln x\right)$$

Solution.

Because limx→0+lnx=−∞$\underset{x\to 0^{+}}{lim}\ln x=-\infty$ and limx→−∞tan−1x=−π2$\underset{x\to -\infty }{lim}{\tan }^{-1}x=-\frac{\pi }{2}$.

Then limx→0+tan−1(lnx)=−π2.$$\underset{x\to 0^{+}}{lim}{\tan }^{-1}\left(\ln x\right)=-\frac{\pi }{2}.$$

7.7 L’hospital’s Rule

Exercise 7.49 Evaluate the limit if it exist. Otherwise, explain why.

limx→1x3−2x2+1x3−1$$\underset{x\to 1}{lim}\frac{x^3-2x^2+1}{x^3-1}$$

Solution.

Because limx→1(x3−2x2+1)=0$\underset{x\to 1}{lim}(x^3-2x^2+1)=0$ and limx→1(x3−1)=0${lim}_{x\to 1}(x^3-1)=0$. Moreover, limx→1(x3−2x2+1)′(x3−1)′=limx→13x2−4x3x2=3−43=−13.$$\begin{array}{cc}\underset{x\to 1}{lim}\frac{(x^3-2x^2+1{)}^{\prime }}{(x^3-1{)}^{\prime }}=& \underset{x\to 1}{lim}\frac{3x^2-4x}{3x^2}\\ =& \frac{3-4}{3}=-\frac{1}{3}.\end{array}$$

By L’Hospital’s rule, limx→1x3−2x2+1x3−1=limx→1(x3−2x2+1)′(x3−1)′=−13.$$\underset{x\to 1}{lim}\frac{x^3-2x^2+1}{x^3-1}=\underset{x\to 1}{lim}\frac{(x^3-2x^2+1{)}^{\prime }}{(x^3-1{)}^{\prime }}=-\frac{1}{3}.$$

Exercise 7.50 Evaluate the limit if it exist. Otherwise, explain why.

limx→0sin4xtan5x$$\underset{x\to 0}{lim}\frac{\sin 4x}{\tan 5x}$$

Solution.

Since limx→0sin4x=0$\underset{x\to 0}{lim}\sin 4x=0$ and limx→0cos4x=0$\underset{x\to 0}{lim}\cos 4x=0$, and limx→0(sin4x)′(tan5x)′=limx→04cos4x5sec25x=45.$$\underset{x\to 0}{lim}\frac{(\sin 4x{)}^{\prime }}{(\tan 5x{)}^{\prime }}=\underset{x\to 0}{lim}\frac{4\cos 4x}{5{\sec }^{2}5x}=\frac{4}{5}.$$

By L’Hospital’s rule, limx→0sin4xtan5x=limx→0(sin4x)′(tan5x)′=45.$$\underset{x\to 0}{lim}\frac{\sin 4x}{\tan 5x}=\underset{x\to 0}{lim}\frac{(\sin 4x{)}^{\prime }}{(\tan 5x{)}^{\prime }}=\frac{4}{5}.$$

Exercise 7.51 Evaluate the limit if it exist. Otherwise, explain why.

limθ→π/21−sinθ1+cos2θ$$\underset{\theta \to \pi /2}{lim}\frac{1-\sin \theta }{1+\cos 2\theta }$$

Solution.

We will apply L’Hospital’s rule twice.

limθ→π/21−sinθ1+cos2θ=limθ→π/2−cosθ−2sin2θ=limθ→π/2−sinθ4cos2θ=−1−4=14.$$\begin{array}{cc}\underset{\theta \to \pi /2}{lim}\frac{1-\sin \theta }{1+\cos 2\theta }=& \underset{\theta \to \pi /2}{lim}\frac{-\cos \theta }{-2\sin 2\theta }\\ =& \underset{\theta \to \pi /2}{lim}\frac{-\sin \theta }{4\cos 2\theta }\\ =& \frac{-1}{-4}=\frac{1}{4}.\end{array}$$

Exercise 7.52 Evaluate the limit if it exist. Otherwise, explain why.

limx→0sin−1xx$$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}$$

Solution.

We may apply L’Hospital’s rule.

limx→0sin−1xx=limθ→01√1−x2=1.$$\begin{array}{cc}\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}=& \underset{\theta \to 0}{lim}\frac{1}{\sqrt{1-x^2}}\\ & =1.\end{array}$$

Exercise 7.53 Evaluate the limit if it exist. Otherwise, explain why.

limx→∞xsin(π/x)$$\underset{x\to \infty }{lim}x\sin \left(\pi /x\right)$$

Solution.

First write the expression as a quotient and then apply L’Hospital’s rule.

limx→∞xsin(π/x)=limx→∞sin(π/x)1/x=limx→∞cos(π/x)⋅(π⋅(1/x)′)(1/x)′=limx→∞πcos(π/x)=π.$$\begin{array}{cc}\underset{x\to \infty }{lim}x\sin \left(\pi /x\right)=& \underset{x\to \infty }{lim}\frac{\sin \left(\pi /x\right)}{1/x}\\ =& \underset{x\to \infty }{lim}\frac{\cos \left(\pi /x\right)\cdot (\pi \cdot (1/x{)}^{\prime })}{(1/x{)}^{\prime }}\\ =& \underset{x\to \infty }{lim}\pi \cos \left(\pi /x\right)=\pi .\end{array}$$

Exercise 7.54 Evaluate the limit if it exist. Otherwise, explain why.

limx→0+(1x−1ex−1)$$\underset{x\to 0^{+}}{lim}(\frac{1}{x}-\frac{1}{e^x-1})$$

Solution.

First write the expression as a quotient and then apply L’Hospital’s rule twice.

limx→0+(1x−1ex−1)=limx→0+ex−x−1x(ex−1)=limx→0+ex−1(ex−1)+xex=limx→0+ex2ex+xex=12.$$\begin{array}{cc}\underset{x\to 0^{+}}{lim}(\frac{1}{x}-\frac{1}{e^x-1})=& \underset{x\to 0^{+}}{lim}\frac{e^x-x-1}{x(e^x-1)}\\ =& \underset{x\to 0^{+}}{lim}\frac{e^x-1}{(e^x-1)+x{e}^x}\\ =& \underset{x\to 0^{+}}{lim}\frac{e^x}{2e^x+x{e}^x}=\frac{1}{2}.\end{array}$$

Exercise 7.55 Evaluate the limit if it exist. Otherwise, explain why.

limx→∞(x−lnx)$$\underset{x\to \infty }{lim}(x-\ln x)$$

Solution.

An estimate using the identity ln(ex)=x$\ln \left(e^x\right)=x$ shows that the limit is ∞$\infty$. To verify that, we factor out x$x$. Note that limx→∞lnxx=limx→∞1x=0.$$\underset{x\to \infty }{lim}\frac{\ln x}{x}=\underset{x\to \infty }{lim}\frac{1}{x}=0.$$

Then limx→∞(x−lnx)=limx→∞x(1−lnxx)=∞.$$\begin{array}{cc}\underset{x\to \infty }{lim}(x-\ln x)=& \underset{x\to \infty }{lim}x(1-\frac{\ln x}{x})\\ =& \infty .\end{array}$$

Exercise 7.56 Find the limit

limx→∞x1/x$$\underset{x\to \infty }{lim}{x}^{1/x}$$

Solution.

Using the identity ln(ex)=x$\ln \left(e^x\right)=x$, we may rewrite x1/x$x^{1/x}$ as x1/x=elnxx.$$x^{1/x}=e^{\frac{\ln x}{x}}.$$

Since limx→∞lnxx=limx→∞1x=0.$$\underset{x\to \infty }{lim}\frac{\ln x}{x}=\underset{x\to \infty }{lim}\frac{1}{x}=0.$$

Then limx→∞x1/x=elimx→∞lnxx=e0=1.$$\underset{x\to \infty }{lim}{x}^{1/x}=e^{\underset{x\to \infty }{lim}\frac{\ln x}{x}}=e^0=1.$$