Answer:
[tex]\displaystyle g'(10)=\frac{1}{-13}=-\frac{1}{13}[/tex]
Step-by-step explanation:
The Derivative of the Inverse Function
Let f(x) be a real invertible function, and g(x) the inverse function of f(x), i.e.,:
[tex]g(x)=f^{-1}(x)[/tex]
We can calculate the derivative of the inverse function even if we don't have the inverse function explicitly computed. We use the formula:
[tex]\displaystyle g'(x)=\frac{1}{f'(g(x))}[/tex]
We need to find the value of g'(10) when:
[tex]f(x)=-x^3-x[/tex]
Substituting:
[tex]\displaystyle g'(10)=\frac{1}{f'(g(10))}[/tex]
We don't have the value of g(10) but we can guess its value since the inverse functions f and g satisfy:
if y=f(x), then g(y)=x, thus we need to find a value of x that produces a value of f(x)=10.
We can easily find that x=-2:
[tex]f(-2)=-(-2)^3-(-2)=8+2=10.[/tex]
Thus, g(10)=-2
Now we find:
[tex]f'(x)=-3x^2-1[/tex]
[tex]f'(-2)=-3(-2)^2-1[/tex]
[tex]f'(-2)=-3*4-1=-13[/tex]
Thus, finally:
[tex]\displaystyle g'(10)=\frac{1}{-13}=-\frac{1}{13}[/tex]
