The correct option is D f′(x) changes its sign twice as x varies from (−∞,∞)
For x<0
f(x)=x33−4x
Hence
f′(x)=x2−4
Now for monotonically increasing function
f′(x)>0
Or
xϵ(−∞,−2)∪(2,∞). However domain is x<0.
Hence xϵ(−∞.−2).
For
0<x<1
f(x)=x3.
f′(x)=3x2
Now for monotonically increasing function f′(x)>0
Or
x>0.
Hence f(x) is an increasing function in (0,1).
For x>1
f(x)=√x
f′(x)=12√2
for monotonically increasing function f′(x)>0
Hence
x>0.
Thus summing up, we get f(x) as an increasing function in the interval of
(−∞,−2)∪(0,∞).
Hence it changes sign twice, once at x=2 and another at x=0.