If f(x)=|x|3, show that f"(x)exists for all real x, and find it.
If f(x)=|x|3, show that f"(x)exists for all real x, and find it.
Solution and Explanation
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 \(f'(x)=\left\{\begin{matrix} 6x,&if\,x\geq 0 \\ -6x,&if\,x<0 \end{matrix}\right.\)
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.