Let $f(x)=6x^3-11x^2-2$, $x\geqslant 1. 5$. Find $\frac{\text{d}f^{-1}}{\text{d}x}$ at the point $x=1517=f(7)$.
I thought you should first find the inverse and then differentiate but from looking at the inverse, that's clearly not what I should do. Any ideas? Also, why did they give me x>=1.5?
On
Hint: Indeed, you don't need to know the inverse function of $f$. Note that by inverse function theorem, you can find $$\frac{df^{-1}}{dx}(b)=(f^{-1})'(b)=\frac{1}{f'(a)}$$ where $b=f(a)$ and only you need show that $f'(a)\not=0$ and then you need to find $\frac{1}{f'(a)}$.
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Just to show you what you did avoid to face.
If $y>0$ the inverse is given by $$x=\frac{11}{18} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{486y+2303}{1331}\right)\right)\right)$$ $$\frac {dx}{dy}=\frac{11 \sinh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{486 y+2303}{1331}\right)\right)}{\sqrt{729 y^2+6909 y+10902}}$$ If $x=1517$, $y=20921018297$ making at this point $$\frac {dx}{dy}=\frac{11 \sinh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{10167614894645}{1331}\right)\right)}{20694914 \sqrt{745016541}}$$ while,as given in the answers, $$\frac {dy}{dx}=18 x^2-22 x$$ which is for $x=1517$ $$\frac {dy}{dx}=41389828 \implies \frac {dx}{dy}=\frac 1 {41389828}$$
Which one do you prefer ?
They gave you $x\ge 1.5$ because the function is invertible in the interval $[1.5,+\infty)$ as you can see in the graph below the function is invertible in the intervals $(-\infty,0]$, $\left[0,\frac{11}{9}\right]$ and $\left[\frac{11}{9},+\infty\right)$. The values are the $x$ of the critical points of the function.
$f'(x)=18 x^2-22 x$ and $f'(7)=728$ thus the derivative of the inverse is $\left(f^{-1}\right)'(1517)=\frac{1}{f'(7)}=\frac{1}{728}\approx 0.0014$. The value is so small because $f(x)$ grows very fast around $x=7$, therefore its inverse grows very slowly.