Let f(x) and g(x) be twice differentiable functions satisfying f''(x)=g''(x) for all xR, w…

Question

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x)=g''(x)$ for all $x\in\mathbb{R}$, with $f'(1)=4$, $2g'(1)=4$, $g(2)=9$, and $3f(2)=9$. Then $f(25)-g(25)$ is equal to:

Accepted Answer

Let $h(x)=f(x)-g(x)$. Then $h''(x)=f''(x)-g''(x)=0$ for all $x$, so $h'(x)$ is constant, say $h'(x)=c_1$.

From the given data, $f'(1)=4$ and $g'(1)=2$ (since $2g'(1)=4$), so
$$c_1=h'(1)=f'(1)-g'(1)=4-2=2 \implies h'(x)=2 \text{ for all }x$$

Integrating, $h(x)=2x+c_2$. From the given data, $g(2)=9$ and $f(2)=3$ (since $3f(2)=9$), so
$$h(2)=f(2)-g(2)=3-9=-6=2(2)+c_2\implies c_2=-10$$

$$h(x)=2x-10 \implies h(25)=50-10=40$$

$f(25)-g(25)=40$

Solutions · 1