Answer:
g'(10) = [tex]\frac{-1}{16}[/tex]
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x (Identity)
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = [tex]\frac{1}{f'(x)}[/tex] --equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = [tex]\frac{4}{x}[/tex] + 2
[tex]\frac{4}{x}[/tex] = 8
x = [tex]\frac{1}{2}[/tex]
Since f(x) = [tex]\frac{4}{x}[/tex] + 2
f'(x) = -[tex]\frac{4}{x^{2} }[/tex]
f'([tex]\frac{1}{2}[/tex]) = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -[tex]\frac{1}{16}[/tex]
