f(x)=4+x^2+tan(pi*x/3), -1
find (f-1)'(a)
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Note that f(x) is increasing on -1 < x < 1. Thus, an inverse will exist on this interval.
Then, since f(0) = 4, then by the definition of the inverse function:
f^(-1)(4) = 0.
Now, since f(x) and f^(-1)(x) satisfy f[f^(-1)(x)] = x, differentiating this using the Chain Rule gives:
f^(-1)'(x) * f'[f^(-1)(x)] = 1 ==> f^(-1)'(x) = 1/f'[f^(-1)(x)].
At x = 4:
f^(-1)'(4) = 1/f'[f^(-1)(4)]
= 1/f'(0), since f^(-1)(4) = 0
= 1/[2(0) + (π/3)(1)], since f'(x) = 2x + (π/3)sec^2(πx/3)
= 3/π.
I hope this helps!
Then, since f(0) = 4, then by the definition of the inverse function:
f^(-1)(4) = 0.
Now, since f(x) and f^(-1)(x) satisfy f[f^(-1)(x)] = x, differentiating this using the Chain Rule gives:
f^(-1)'(x) * f'[f^(-1)(x)] = 1 ==> f^(-1)'(x) = 1/f'[f^(-1)(x)].
At x = 4:
f^(-1)'(4) = 1/f'[f^(-1)(4)]
= 1/f'(0), since f^(-1)(4) = 0
= 1/[2(0) + (π/3)(1)], since f'(x) = 2x + (π/3)sec^2(πx/3)
= 3/π.
I hope this helps!