f(x) and g(x) are two differentiable functions on [0,2] such that f''(x)-g''(x)=0, f'(1)=2…

Question

$f(x)$ and $g(x)$ are two differentiable functions on $[0,2]$ such that $f''(x)-g''(x)=0$, $f'(1)=2$, $g'(1)=4$, $f(2)=3$, $g(2)=9$ then $f(x)-g(x)$ at $x=3/2$ is:

*(reading the garbled equalities as: $f'(1)=2$, $g'(1)=4$, $f(2)=3$, $g(2)=9$ --- matching the worked solution's numbers)*

Accepted Answer

$f''(x)-g''(x)=0$. Integrating: $f'(x)-g'(x)=c$; using $f'(1)-g'(1)=c$: from source values $4-2=c \Rightarrow c=2$

$	herefore f'(x)-g'(x)=2$

Integrating again: $f(x)-g(x)=2x+c_1$

Using $f(2)-g(2)=4+c_1$: $9-3=4+c_1 \Rightarrow c_1=2$

$	herefore f(x)-g(x)=2x+2$

At $x=3/2$: $f(x)-g(x) = 3+2 = 5$

Solutions · 1