Tricky Corner!

Calculus Level 1

Above shows the graph of $f(x) = | x|$ , what is the value of $f'(0)$ ?

0 2 -1 Not Differentiable at

x=0

5 solutions

Michael Fuller
Jun 13, 2015

$\large {f(x)=|x|=\sqrt{{x}^{2}}={({x}^{2})}^{\frac{1}{2}} \\ f'(x)=\frac{1}{2}{({x}^{2})}^{-\frac{1}{2}}\times 2x =\frac{x}{|x|} \\ f'(0)=\frac{0}{|0|}}$

$\large \therefore \color{#20A900}{\boxed{Not \quad differentiable \quad at \quad x=0}}$

Another way is to find the one-sided derivatives of $f(x)$ at $x=0$ . By definition of the absolute value function, we have,

$\lim_{x\to 0^-}f^\prime(x)=(-1)\quad\textrm{and}\quad\lim_{x\to 0^+}f^\prime(x)=1$

Since the left and right derivatives are unequal, we conclude that $f^\prime(0)$ doesn't exist, ergo, $f(x)$ is not differentiable at $x=0$ .

Prasun Biswas - 6 years ago

we also can draw infinite number of tangent lines at that point so we cannot determine what is the derivative at x=0

Mohamed Ameén - 5 years, 12 months ago

Vikas Khirwadkar
Jun 13, 2015

When there is a sharp edge, function cannot be differentiable. here at point 0 there is a sharp edge hence f'(0) is not differentiable.

I think sharp is not so rigorous term. Consider $f(x) = 15x^{4/3}-12x^{5/3}$ and see whether it's sharp on x = 0

Bostang Palaguna - 7 months ago

So I think proofing that the left limit and right limit of the derrivative not the same is the most convincing way yo answer this

Bostang Palaguna - 7 months ago

Sam Javier
Jul 23, 2015

Recall that a function is not differentiable at some point x if it has a sharp edge at that particular point.

Erica Phillips
Apr 12, 2018

as u can see the no. of slopes we can draw through the point 0 hence the slope is not fixed as multiple lines like the above can be drawn!!!

Ameya Bhamare
Aug 4, 2015

The function has sharp point on x=0 making it non deferenciable at 0

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