Question

The function $f\left(x\right)=1–x^3–x^5$ is decreasing for

• A. $1\le x\le 5$
• B. $x\le 1$
• C. $x\ge 1$
• D. All values of $x$

Answer 1

The correct option is D

All values of $x$

Explanation for the correct option:

Given function $f\left(x\right)=1–x^3–x^5$

So $f\text{'}\left(x\right)=-3x^2-5x^4$

The critical point of $f\left(x\right)$ are the values of $x$ for which $f\text{'}\left(x\right)=0$.
$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \Rightarrow -3x^2-5x^4=0 \\ \Rightarrow -x^2\left(3+5x^2\right)=0 \\ \Rightarrow x=0,\sqrt{\frac{-3}{5}} \end{gathered}$$

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

And, $f\text{'}\left(x\right)=-3x^2-5x^4<0,\forall x\in R.$

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

Hence, $\mathrm{Option}\left(\mathrm{D}\right)$ is the correct answer