Question

let f and g be differentiable functions with the following properties: (i) g(x)>0 for all x (ii) f(0)=1 if h(x)=f(x)g(x) and h(x)=f(x)g(x), then f(x)= a f(x) b g(x) c e^x d 0

Explanation:

Step1: Apply product - rule of differentiation

The product - rule states that if $h(x)=f(x)g(x)$, then $h^{\prime}(x)=f(x)g^{\prime}(x)+f^{\prime}(x)g(x)$. Given that $h^{\prime}(x)=f(x)g^{\prime}(x)$, we have $f(x)g^{\prime}(x)+f^{\prime}(x)g(x)=f(x)g^{\prime}(x)$.

Step2: Simplify the equation

Subtract $f(x)g^{\prime}(x)$ from both sides of the equation $f(x)g^{\prime}(x)+f^{\prime}(x)g(x)=f(x)g^{\prime}(x)$. We get $f^{\prime}(x)g(x)=0$. Since $g(x)>0$ for all $x$, then $f^{\prime}(x) = 0$ for all $x$.

Step3: Find the function $f(x)$

A function whose derivative is zero for all $x$ is a constant function. Let $f(x)=C$. Given that $f(0) = 1$, substituting $x = 0$ into $f(x)=C$, we find that $C = 1$. So $f(x)=1$.

Answer:

A. $f(x)$