Use the function f and the given real number a to find (f^-1)'(a)

f(x)=x^3+8x-3 a=6
(f^-1)'(6)=

Thank you.

6 answers

Do you mean
f(x) = x^2 + 8 x - 3 ??
then to find f'-1
x = y^2 + 8 y -3

put in x = 6

y^2 + 8 y - 9 = 0

(y-1)(y+9) = 0
y = 1 or -9
No it is x^3
did not see you wanted derivative either

x = y^3 + 8 y - 3

dx = 3 y^2 dy + 8 dy

dy/dx = 1/[3 y^2 + 8]

if y = 6

1/[116]
1/116 is not correct
The answer is 1/11
f = x^3+8x-3
f' = 3x^2+8
f'(6) = 116

dx/df = 1/(df/dx) = 1/116

1/11 must be a typo.

http://www.sosmath.com/calculus/diff/der08/der08.html
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