Question

If $f\left(x\right)$ is a function such that $f^{\prime }\left(x\right)=(x-1{)}^2(4-x)$, then

• A. $f\left(0\right)=0$
• B. $f\left(x\right)$ is increasing in $(0,3)$
• C. $x=4$ is a critical point of $f\left(x\right)$
• D. $f\left(x\right)$ is decreasing in (3, 5)

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(x\right)$ is increasing in $(0,3)$
C $x=4$ is a critical point of $f\left(x\right)$

Sign scheme of $f^{\prime }\left(x\right)$ can be drawn as


In the interval $(0,3)$
$f^{\prime }\left(x\right)>0$
$\therefore f\left(x\right)$ is increasing in $(0,3)$

As$f^{\prime }\left(4\right)=0$
So $x=4$ is a critical point

Integrating $f^{\prime }\left(x\right)$ w.r.t $x$
$f\left(x\right)=-\frac{x^4}{4}+2x^3-\frac{9x^2}{2}+4x+c$,
where $c$ is a constant
So we can't determine whether $f\left(0\right)=0$

In the interval $(3,5),f^{\prime }\left(x\right)$ changes it's sign at $x=4$, so it's not decreasing in the given interval.