Using the Definition to Compute the Derivative - Exercise 2

Exercise 2. Discuss the differentiability of

$$f(x) = \vert x^2 - x\vert\;\cdot$$

Answer. May be the scariest thing about this function is the absolute value. So the best thing to do is to look for ways to remove it. Therefore we are led to find out when x² - x is positive or negative. We get

$$f(x) = \left\{\begin{array}{rll} x^2 - x &\mbox{if $x \leq 0$... ... 1$}\\ x^2 - x &\mbox{if $x \geq 1$}\;.\\ \end{array} \right.$$

Clearly the derivative exists at every point, except maybe at 0 and 1. Let us discuss these two points. Let us start with 0. We have

$$f'(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \rightarrow 0} \frac{f(x)}{x}\;\cdot$$

Since the function is defined differently from the left and the right of 0, then we have to consider the limits to the left and to the right at 0. We have

$$\lim_{x \rightarrow 0-} \frac{f(x)}{x} = \lim_{x \rightarrow 0-} \frac{x^2-x}{x} = -1$$

and

$$\lim_{x \rightarrow 0+} \frac{f(x)}{x} = \lim_{x \rightarrow 0+} \frac{-(x^2-x)}{x} = 1\;.$$

This implies that f'(0) does not exist. Similar computations will also give

$$\lim_{x \rightarrow 1-} \frac{f(x)-f(1)}{x-1} = \lim_{x \rightarrow 1-} \frac{-(x^2-x)}{x-1} = -1$$

and

$$\lim_{x \rightarrow 1+} \frac{f(x)-f(1)}{x-1} = \lim_{x \rightarrow 1+} \frac{x^2-x}{x-1} = 1$$

which implies that f'(1) does not exist.

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