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If f(x)=x^3+4x+6, find (f−1)′(6)
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its actaully( f^-1)'(6)
i meant to correct that. sorry
\[(f^{-1})'(6)=\frac{1}{f'(f^{-1}(6))}=\frac{1}{f'(0)} \text{ since } f(0)=6 => f^{-1}(6)=0\]
0 doesnt work
Can i have some help?
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\[f'(x)=3x^2+4 =>f'(0)=3(0)^2+4=4 => (f^{-1})'(6)=\frac{1}{4}\]
Oh, he meant inverse. I'm sorry. I'll delete my answer.
1/4 works thanks myinaya
I know it works lol
Oh you thought I was saying the answer is 0? Do you understand the steps?
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