I have been given the function: $f(x)=x^5+3x-2$. How do I find the inverse of this. When I switch x and y, it just goes downhill from there and I'm not even sure what I'm supposed to do after I figure out the inverse. Is there any tricks to find the inverse with the powers??
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I don't think that there is a closed form for the inverse, if that's what you're after. – flawr Oct 26 '16 at 21:20
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Well I'm not even sure how to find it in the first place – Justin Oct 26 '16 at 21:21
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If you can find one, the approach you mentioned is usually the way to go. But you still might find the derivative by deriving $x = f(f^{-1}(x))$ – flawr Oct 26 '16 at 21:22
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finding the derivative of the inverse function is not as hard as finding the inverse itself... – Will Jagy Oct 26 '16 at 21:23
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Given the title it looks more as if you need the formula for the derivative of the inverse (without knowing the inverse)?
When $y_0=f(x_0)$ and $f'(x_0)\neq 0$ then the $C^1$ function $f$ has a local $C^1$ inverse $g$ defined in a neighborhood $N_y$ of $y_0$ with values in a neighborhood $N_x$ of $x_0$. One has $f(g(y))=y$ for $y\in N_y$ and $g(f(x))=x$ for $x\in N_x$ which by taking derivatives leads to: $$ g'(y_0)=\frac{1}{f'(x_0)} = \frac{1}{f'(g(y_0))} $$ So for example $f(1)=2$, $f'(1)=8$ so the local inverse at $2$ verifies $g'(2)=1/8$. In the present case $f$ is strictly increasing and is bijection onto ${\Bbb R}$ so $g$ is defined on all of ${\Bbb R}$.
H. H. Rugh
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You can verify that $f$ is strictly monotonous, so an inverse $g=f^{-1}$ exists. Then we know that $y = f(g(y))$. We can derive this equation and get $$ 1 = f'(g(y)) g'(y)$$
This implies for the point $x=f(x)$ that
$$ g'(y) = g'(f(x)) = \frac{1}{f'(g(y))} = \frac{1}{f'(x)}$$
flawr
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