Question

If $f\text{'}\left(x\right)=g\left(x\right){(x-a)}^2$, where $g\left(a\right)\ne 0$ and $g$ is continuous at $x=a$, then $f\text{'}\left(x\right)$:

• A. $f$ is increasing in the nbd. of $a$ if $g\left(a\right)>0$ and $f$ is decreasing in the nbd. of $a$ if $g\left(a\right)<0$.
• B. $f$ is increasing in the nbd. of $a$ if $g\left(a\right)<0$.
• C. $f$ is decreasing in the nbd. of a if $g\left(a\right)>0$.
• D. $f$ is increasing in the nbd. of $a$ if $g\left(a\right)<0$ and $f$ is decreasing in the nbd. of $a$ if $g\left(a\right)>0$.

Answer 1

The correct option is A

$f$ is increasing in the nbd. of $a$ if $g\left(a\right)>0$ and $f$ is decreasing in the nbd. of $a$ if $g\left(a\right)<0$.

Explanation for the correct option.

It is given that $g\left(a\right)\ne 0$, that means either $g\left(a\right)>0$ or $g\left(a\right)<0$.

Now, if $g\left(a\right)>0$, then $g\left(x\right)>0$ as $g$ is continuous at $x=a$. So, $f\left(x\right)$ is increasing in the nbd. of $a$ if $g\left(a\right)>0$.

If $g\left(a\right)<0$, then $g\left(x\right)<0$ as $g$ is continuous at $x=a$. So, $f\left(x\right)$ is decreasing in the nbd. of $a$ if $g\left(a\right)<0$.

Therefore, the correct answer is option A.