Question

If $f\left(x\right)$ is 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 $f\left(x\right)$

The function $f\left(x\right)$ is increasing for $x=x_1$ if $f^{\prime }\left(x\right)>0$ for $x=x_1$.
Therefore,
$f^{\prime }\left(x\right)>0$
$(x-1{)}^2(4-x)>0$
Since $(x-1{)}^2\ge 0$ for all $x\in R$, hence
$4-x>0$ or $x<4$.
Therefore, $f\left(x\right)$ is increasing for $x<4$.
Also
$f^{\prime }\left(x\right)=0$ implies
$(x-1{)}^2(4-x)=0$
Therefore,
$x=1$ and $x=4$. Hence $x=1$ and $x=4$ are critical points.