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Chapter 4: Problem 29

\(y=f^{-1}(x)\) As $g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^{3}}=-\frac{4}{8}=-\frac{1}{2}$

Short Answer

Expert verified
Question: Given the second derivative of the function g with respect to y, and some information about the first and second derivatives of the function f in terms of x, find the expression for the inverse function y=f^{-1}(x). Solution: While we cannot find the explicit expression for y=f^{-1}(x), we have found that the first derivative of the inverse function is \(f^{-1}(x)^{\prime}=\frac{1}{f'(x)}\), and the second derivative is \(f^{-1}(x)^{\prime\prime}=-\frac{1}{2}\).

Step by step solution

01

Apply the inverse function theorem

To find the expression for the first derivative of the inverse function, we will apply the inverse function theorem: \(\left(f^{-1}(x)\right)^{'}=\frac{1}{f'(x)}\)
02

Determine the equation of the function y=f^{-1}(x)

We are given the following information: \(g^{\prime \prime}(y)=-\frac{f^{\prime\prime}(x)}{\left(f^{\prime}(x)\right)^{3}}=-\frac{4}{8}=-\frac{1}{2}\) Let's first solve for the second derivative of the inverse function, \(f^{-1}(x)^{\prime\prime}\): \(f^{-1}(x)^{\prime\prime} = g^{\prime\prime}(y) = -\frac{1}{2}\) Next, since \(f^{-1}(x)^{\prime} = \frac{1}{f'(x)}\), we can use \(g^{\prime\prime}(y)=-\frac{f^{\prime\prime}(x)}{\left(f^{\prime}(x)\right)^{3}}=-\frac{4}{8}\) to find the relationship between \(f'(x)\) and \(f''(x)\): \(-\frac{f^{\prime\prime}(x)}{\left(f'(x)\right)^3} = -\frac{4}{8}\) Multiplying both sides by \(-1\), we have: \(\frac{f^{\prime\prime}(x)}{\left(f'(x)\right)^3} = \frac{1}{2}\) Now multiply both sides by \(\left(f'(x)\right)^3\) to move the denominator to the other side: \(f^{\prime\prime}(x) = \frac{1}{2}\left(f'(x)\right)^3\)
03

Solve for f'(x)

To further find the relationship between \(f'(x)\) and \(x\), substitute the value of \(f^{-1}(x)^{\prime}\) obtained in step 1 into the expression for \(f'(x)\): \(f'(x)=\frac{1}{f^{-1}(x)^{\prime}}=\frac{1}{\frac{1}{f'(x)}}\) Let's square both sides of the expression for \(f^{\prime\prime}(x)\): \(\left(f^{\prime\prime}(x)\right)^2=\left(\frac{1}{2}\left(f'(x)\right)^3\right)^2\) \(\left(f^{\prime\prime}(x)\right)^2=\left(\frac{1}{4}\left(f'(x)\right)^6\right)\) Now, notice that we can't solve for \(f'(x)\) explicitly, but the information given in the problem is sufficient to conclude that the function y=f^{-1}(x) exists.
04

Conclude the solution

We have found that the first derivative of the inverse function is \(f^{-1}(x)^{\prime}=\frac{1}{f'(x)}\), and the second derivative is \(f^{-1}(x)^{\prime\prime}=-\frac{1}{2}\). The information given does not allow us to find the explicit expression for \(f'(x)\), nor the function y=f^{-1}(x) itself. However, we have found the information regarding the derivatives of the inverse function, which is the main objective of this exercise.

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Most popular questions from this chapter

\(f(x)+2 f(-x)=-\sin x\) Put \(x=-x\) \(f(-x)+2 f(x)=-\sin x\) Solve, (I) \& (II), we get \(f(x)=-\sin x\) \(f^{\prime}(x)=-\cos x\) \(\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\)

\(a x^{2}+b y^{2}+2 h x y=1\) \(2 a x+2 b y y^{\prime}+2 h y+2 h x y^{\prime}=0\) \(y^{\prime}=\frac{-(a x+h y)}{h x+b y}\) Differentiating again eq (I) $a+b\left(y^{\prime}\right)^{2}+b y y^{\prime \prime}+h y^{\prime}+h x y^{\prime \prime}+h y^{\prime}=0$ $\Rightarrow y^{\prime \prime}=\frac{-\left(a+b\left(y^{\prime}\right)^{2}+2 h y^{\prime}\right)}{(h x+b y)}$ $=\frac{-1}{(h x+b y)^{3}}\left[\begin{array}{l}a(h x+b y)^{2}+b(a x+h y)^{2} \\\ -2 h(a x+h y)(h x+b y)\end{array}\right]$ $=\frac{-1}{(h x+b y)^{3}}\left[\begin{array}{l}a h^{2} x^{2}+a b^{2} y^{2}+2 a b h x y+b a^{2} x^{2}+b h^{2} y^{2} \\ +2 a b h x y-2 a h^{2} x^{2}-2 a b h x y-2 h^{3} x y-2 h^{2} b y^{2}\end{array}\right]$ $=\frac{-1}{(h x+b y)^{3}}\left[a b^{2} y^{2}+b a^{2} x^{2}-a h^{2} x^{2}-h^{2} b y^{2}-2 h^{3} x y+2 a b h x y\right]$ $\quad=\frac{-1}{(h x+b y)^{3}}\left[a b\left(b y^{2}+a x^{2}+2 h x y\right)-h^{2}\left(a x^{2}+b y^{2}+2 h x y\right)\right]$ \(\quad=\frac{h^{2}-a b}{(h x+b y)^{3}}\)

$\begin{aligned} &f(x)=\left|\begin{array}{lll} \cos x & \sin x & \cos x \\ \cos 2 x & \sin 2 x & 2 \cos 2 x \\ \cos 3 x & \sin 3 x & 2 \cos 3 x \end{array}\right| \\ &f^{\prime}\left(\frac{\pi}{2}\right)=\left|\begin{array}{ccc} -1 & 1 & 0 \\ 0 & 0 & -2 \\ +3 & -1 & 0 \end{array}\right| \\ &+\left|\begin{array}{ccc} 0 & 0 & 0 \\ -1 & -2 & -2 \\ 0 & 0 & 0 \end{array}\right|+\left|\begin{array}{rrr} 0 & 1 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 6 \end{array}\right| \\ &=1 \end{aligned}$

\(f(x)=\frac{x}{1+e^{1 / x}}\) \(f(x)\left(1+e^{1 / x}\right)-x=0\) \(f^{\prime}(x)\left(1+e^{1 / x}\right)-\frac{f(x) e^{l / x}}{x^{2}}-1=0\) \(x^{2} f^{\prime}(x) \frac{(x)}{f(x)}-f(x) e^{1 / x}-x^{2}=0\)

$\begin{aligned} &f(x)=\sqrt{1-\sin 2 x}=\sin x-\cos x \mid \\ &\text { For } x \in(0, \pi / 4) \\ &f(x)=\cos x-\sin x \\ &f^{\prime}(x)=-(\sin x+\cos x) \\ &\text { For } x \in(\pi / 4, \pi / 2) \\ &f(x)=\sin x-\cos x \\ &f^{\prime}(x)=\cos x+\sin x \end{aligned}$
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