Does f^-1(x) even exist?
f'(x) = 5x^4+6x^2+1
That is always positive, which means that the graph of f(x) does not change direction, so it passes the horizontal-line test.
where is f(x) = 3?
x^5+2x^3+x-1 = 3
x^5+2x^3+x-4 = 0
a little synthetic division shows that x=1 is a solution.
So, f^-1(3) = 1
because f(1) = 3
Google inverse function derivative to find that the slopes of f(x) and f^-1(x) are reciprocals.
f'(3) = 556, so (f^-1)'(3) = 1/556
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f(x) = cosx+3x
f'(x) = 1-sinx
Since f' does not change sign, f has an inverse. Use google for discussions on inverse differentiability.
f(0) = 1, so f^-1(1) = 0.
1. Let f(x)=x^5 + 2x^3 + x - 1
Find f^-1(3) and (f^-1)'(3)?
I have zero idea how to find the inverse of this function at a point 3, and how to take derivative of an inverse.
2.Let f(x)=cosx + 3x
Show that f(x) is a differentiable inverse and find f^-1(1) and (f^-1)'(1).
Inverse again. The answer key says yes, it is differentiable, f^-1(1)=0 and (f^-1)'(1)=1/3.
Thank you so much, I really tried swapping x and y values but I didn't get anywhere with that. Online derivative solver didn't work either. PLease help!
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