Answer:
The value of g'(10)=[tex]\frac{(-1)}{16}[/tex]
Step-by-step explanation:
Given function is f(x)=[tex]\frac{4}{x} + 2[/tex]
Take f(x)=y
y=[tex]\frac{4}{x} + 2[/tex]
Subtract 2 from both side.
y-2=[tex]\frac{4}{x}[/tex]
x=[tex]\frac{4}{y-2}[/tex]
The inverse of f(x) is written as [tex] \frac{4}{y-2}[/tex]
It is said as g is inverse of f
g(y)=\frac{4}{y-2}[/tex]
g(y)=[tex]\frac{4}{y-2}[/tex]
g(10)=[tex]\frac{4}{10-2}[/tex]
g(10)=[tex]\frac{1}{2}[/tex]
Differentiating both sides we get,
g'(y)=[tex]\frac{4(-1)}{(y-2)^{2}}+0[/tex]
g'(y)=[tex]\frac{(-4)}{(y-2)^{2}}[/tex]
To find g'(10)=[tex]\frac{(-4)}{(10-2)^{2}}[/tex]
g'(10)=[tex]\frac{(-1)}{16}[/tex]
