Question

If $f\left(x\right)=\left\{\begin{array}{l}0x=0\\ x-3x>0\end{array}\right.$, then the function $f\left(x\right)$ is

• A. increasing when $x>0$
• B. strictly increasing when $x>0$
• C. strictly increasing at $x=0$
• D. not continuous at $x=0$ and so it is not increasing when $x>0$

Answer 1

The correct option is B

strictly increasing when $x>0$

Explanation for the correct option.

Find the nature of the given function.

For $x>0$ the function is defined as $f\left(x\right)=x-3$, now the derivative of the function is $f\text{'}\left(x\right)=1$.

So it can be seen that the function in the interval $x>0$ is such that $f\text{'}\left(x\right)>0$, so the function is strictly increasing in this interval.

The function $f\left(x\right)$ is strictly increasing when $x>0$.

Hence, the correct option is B.