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.
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
To find (f^-1)'(a), we need to find the derivative of the inverse function of f evaluated at a.
First, let's find the inverse of f.
Step 1: Set f(x) = a and solve for x:
x^3 + 8x - 3 = 6
Step 2: Simplify the equation by moving all terms to one side:
x^3 + 8x - 3 - 6 = 0
x^3 + 8x - 9 = 0
Step 3: Solve this cubic equation to find the possible values of x.
In this case, finding the exact solution of the equation is quite difficult. Hence, we can either use numerical methods or approximations to find the value of x. For instance, we can use numerical techniques like Newton's method or the bisection method to find approximate values of x that satisfy the equation.
Let's assume one of the values for x that satisfies the equation is x = p.
Step 4: Compute the derivative of f(x):
f'(x) = 3x^2 + 8
Step 5: Evaluate the derivative at x = p (the approximate value found in Step 3):
f'(p) = 3p^2 + 8
Finally, we have the value of (f^-1)'(a) as f'(p). So, (f^-1)'(6) = 3p^2 + 8, where p is the approximate value of x that satisfies the equation x^3 + 8x - 9 = 0.
Note: The exact value of p depends on the numerical method used to approximate the solution of the cubic equation.