Question

If f(x)=|x|³, show that f"(x)exists for all real x, and find it.

Answer 1

It is known that,\(|x|=\left\{\begin{matrix}  x,&if\,x\geq 0 \\   -x,&if\,x<0  \end{matrix}\right.\)

if x<0 Therefore, when x≥0, 

f(x)=|x|³=x³ In this case,

f'(x)=3x2 and hence,f""(x)=6x 

when x<0,f(x)=|x|³=(-x)³=-x³ In this case,f'(x)=-3x2 and hence,f""(x)=-6x Thus,for f(x)=|x|³

f""(x)exists for all real x and is given by \(f'(x)=\left\{\begin{matrix}  6x,&if\,x\geq 0 \\   -6x,&if\,x<0  \end{matrix}\right.\)