Question

Which of the following statement is true for the function
$f\left(x\right)=\sqrt{x}x\ge 1=x^{3}0\le x\le 1=\frac{x^3}{3}-4xx<0$

• A. It is monotonic increasing for all $xϵR$
• B. $f^{\prime }\left(x\right)$ fails to exist for $3$ distinct real values of $x$
• C. $f^{\prime }\left(x\right)$ changes its sign twice as $x$ varies from $(-\infty ,\infty )$
• D. function attains its extreme values at $x_1$ & $x_2$, such that $x_1,x_2>0$

Answer 1

Answer: (C)

$f^{\prime }\left(x\right)$ changes its sign twice as $x$ varies from $(-\infty ,\infty )$

For $x<0$
$f\left(x\right)=\frac{x^3}{3}-4x$
Hence
$f^{\prime }\left(x\right)=x^2-4$
Now for monotonically increasing function
$f^{\prime }\left(x\right)>0$
Or
$xϵ(-\infty ,-2)\cup (2,\infty )$. However domain is $x<0$.
Hence $xϵ(-\infty .-2)$.
For
$0<x<1$
$f\left(x\right)=x^3$.
$f^{\prime }\left(x\right)=3x^2$
Now for monotonically increasing function $f^{\prime }\left(x\right)>0$
Or
$x>0$.
Hence $f\left(x\right)$ is an increasing function in (0,1).
For $x>1$
$f\left(x\right)=\sqrt{x}$
$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{2}}$
for monotonically increasing function $f^{\prime }\left(x\right)>0$
Hence
$x>0$.
Thus summing up, we get f(x) as an increasing function in the interval of
$(-\infty ,-2)\cup (0,\infty )$.
Hence it changes sign twice, once at $x=2$ and another at $x=0$.