Using the Definition to Compute the Derivative - Exercise 3


Exercise 3. We say that the graph of f(x) has a cusp at (a,f(a)), if f(x) is continuous at a and if the following two conditions hold:

1.
$f'(x) \rightarrow +\infty$ as $x \rightarrow a$ from one side (left or right);
2.
$f'(x) \rightarrow -\infty$ as $x \rightarrow a$ from the other side.

Determine whether f(x) = x4/3 and g(x) = x3/5 have a cusp at (0,0).

Answer. For $x \neq 0$, we have

\begin{displaymath}f'(x) = \frac{4}{3} x^{4/3 - 1} = \frac{4}{3} x^{1/3}.\end{displaymath}

So

\begin{displaymath}\lim_{x \rightarrow 0} f'(x) = 0.\end{displaymath}

f(x) does not have a cusp at 0. In fact, the graph has a horizontal tangent line at (0,0).

For the function g(x), we have

\begin{displaymath}g'(x) = \frac{3}{5} x^{3/5 - 1} = \frac{3}{5} x^{-2/5} = \frac{3}{5}\frac{1}{\Big(x^{1/5}\Big)^2}.\end{displaymath}

In this case, we have

\begin{displaymath}\lim_{x \rightarrow 0+} g'(x) = \lim_{x \rightarrow 0-} g'(x) = +\infty .\end{displaymath}

So again (0,0) is not a cusp for g(x). But in this case, the graph has a vertical tangent at this point. Remember that a vertical line does not have a slope. So the derivative of g(x)at 0 does not exist.


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