Inverse Funtions (1 Viewer)

coyazayo · Oct 3, 2016

Hey everyone! Thanks in advance

1. f(x)=x^3-3x

Find the gradient of the tangent to the curve y=f-1(x) at the point (-11/8,1/2)

2. Let g(x)=e^x-1/e^x

Find an expression for y=f-1(x) in terms of x

InteGrand · Oct 3, 2016

coyazayo said:
Hey everyone! Thanks in advance

1. f(x)=x^3-3x

Find the gradient of the tangent to the curve y=f-1(x) at the point (-11/8,1/2)

2. Let g(x)=e^x-1/e^x

Find an expression for y=f-1(x) in terms of x

$\noindent 1) We have $f(x) = x^{3}-3x$. It follows from the \textsl{inverse function theorem} that$

$\begin{align*}\left(f^{-1}\right)^{\prime}\left(-\frac{11}{8}\right) &= \frac{1}{f^{\prime}\left(\frac{1}{2}\right)}\\ &= \frac{1}{3\left(\frac{1}{2}\right)^{2}-3}\\ &= \frac{8}{3 - 24}\\ &= - \frac{8}{21}.\end{align*}$

$\noindent 2) Note that $g$ is a bijection on $\mathbb{R}$ and so is invertible. Write $x = e^{y} - e^{-y}$, where $y = g^{-1}\left(x\right)$ is the inverse function. Then multiplying through by $e^{y}$ (which is never $0$ for real $y$), we have$

$\begin{align*}e^{2y} - xe^{y} - 1 &= 0 \\ \Rightarrow e^{y} &= \frac{x\pm \sqrt{x^{2}+4}}{2}. \end{align*}$

$\noindent Since $e^y$ is positive, we must take the positive square root (the negative one would give a negative answer overall). So $e^{y} = \frac{x+\sqrt{x^{2}+4}}{2}$. Hence $y = g^{-1} (x) = \ln \left(\frac{x+\sqrt{x^{2}+4}}{2}\right)$.$

Inverse Funtions (1 Viewer)

coyazayo

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InteGrand

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