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 the values of $x$

Answer 1

Answer: (D)

All the values of $x$

$F\left(x\right)=1-x^3-x^5$

The critical point of $F\left(x\right)$ are the values of $x$

for which either $f^{\prime }\left(x\right)$ does not exist or $f^{\prime }\left(x\right)=0$

So $f\left(x\right)=1-x^3-x^5$

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

$f^{\prime }\left(x\right)=0$ when $x=0$

$x=0$ is the only critical point because $3+5x^2>0$

for all $x\in R$

f is decreasing on all the x value $(x\in R)$

because $f^{\prime }\left(x\right)$ < 0 for all $x\in R$