Question

6. Let f and g be differentiable functions

Answer 1

Answer:

(b) 1

Step-by-step explanation:

To differentiate $h(x)=f(x)g(x)$ we will need the product rule:

$h'(x)=f'(x)g(x)+f(x)g'(x)$.

We have $h'(x)=f(x)g'(x)$, so the following equation is true by the transitive property:

$f'(x)g(x)+f(x)g'(x)=f(x)g'(x)$

By subtraction property we have:

$f'(x)g(x)=0$

Since $g(x) \neq 0$, then we can divide both sides by $g(x)$:

$f'(x)=\frac{0}{g(x)}$

$f'(x)=0$

This implies $f(x)$ is constant.

So we have that $f(x)=c$ where $c$ is a real number.

Since $f(0)=1$ and $f(0)=c$, then by transitive property $1=c$.

So $f(x)=1$.

Checking:

$h(x)=1 \cdot g(x)$

$h'(x)=g'(x)=1 \cdot g'(x)$

So the following conditions were met.