Question

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

Answer 1

» It is the know that :-

$= > |x| = x \: \: if \: \: x \geqslant 0 \\ \\ = > |x| = - x \: \: if \: \: x \leqslant 0$
_________

Therefore ,

when

$= > x \geqslant 0 \: \\ \\ = > f(x) = { |x| }^{3} = {x}^{3}$
_____________

In this case :-

=> f'(X) = 3x²

and hence,

=> f"(X) = 6x

_____________

when x < 0

$= > f(x) = | {x}^{3} | = - {x}^{3}$
in this case :-

=> f'(X) = -3x²

and hence,

=> f"(X) = -6x

_________________

Thus, for

$= > f(x) = { |x| }^{3}$
» f"(X) exist for all real x and is given by

=> f"(X) = 6x __[if x ≥ 0]

and

=> f"(X) = -6x __[if x ≤ 0]
___________________[AMSWER]

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_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_

Answer 2

modulus is the piece-wise function, we can't differentiate without breaking mode.
we know,
$\bf{|x|=\left\{\begin{array}{ll}x&x\geq0\\-x&x<0\end{array}\right.}$

$\bf{so, f(x)=|x|^3=\left\{\begin{array}{ll}x^3&x\geq0\\-x^3&x<0\end{array}\right.}$

now, when x ≥ 0
f(x) = x³
differentiate f(x) with respect to x,
f'(x) = d(x³)/dx = 3x²
again differentiate with respect to x,
f''(x) = d(3x²)/dx = 3.2x = 6x

when x < 0
f(x) = -x³
differentiate f(x) with respect to x,
f'(x) = - 3x²
differentiate once again,
f"(x) = d(-3x²)/dx = -6x

hence, $\bf{f"(x)=\left\{\begin{array}{ll}6x&x\geq0\\-6x&x<0\end{array}\right.}$