Question

The function $f\left(x\right)=1-x^3$:

• A. increases everwhere
• B. decreases in $(0,\infty )$
• C. increases in $(0,\infty )$
• D. None of the above

Answer 1

The correct option is B

decreases in $(0,\infty )$

Explanation for correct option

Given function $f\left(x\right)=1-x^3$

So $f\text{'}\left(x\right)=-3x^2$

The critical point of $f\left(x\right)$ are the values of $x$ for which $f\text{'}\left(x\right)=0$.
Thus, $f\text{'}\left(x\right)=0$ when $x=0$

So, $x=0$ is the only critical point.

$f\text{'}\left(x\right)=-3x^2<0,\forall x\in R.$

Thus, $f\left(x\right)$ is strictly decreasing for all value of $x$.

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.