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
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
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]
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
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
0 answers