4.5 Derivatives and the shape of a graph  (Page 3/14)

Using the first derivative test

Use the first derivative test to find the location of all local extrema for $f\left(x\right)=5x^{1\text{/}3}-x^{5\text{/}3}.$ Use a graphing utility to confirm your results.

Step 1. The derivative is

$f^{\prime }\left(x\right)=\frac{5}{3}{x}^{\mathrm{-2}\text{/}3}-\frac{5}{3}{x}^{2\text{/}3}=\frac{5}{3x^{2\text{/}3}}-\frac{5x^{2\text{/}3}}{3}=\frac{5-5x^{4\text{/}3}}{3x^{2\text{/}3}}=\frac{5\left(1-x^{4\text{/}3}\right)}{3x^{2\text{/}3}}.$

The derivative $f^{\prime }\left(x\right)=0$ when $1-x^{4\text{/}3}=0.$ Therefore, $f^{\prime }\left(x\right)=0$ at $x=\text{±}1.$ The derivative $f^{\prime }\left(x\right)$ is undefined at $x=0.$ Therefore, we have three critical points: $x=0,$ $x=1,$ and $x=\mathrm{-1}.$ Consequently, divide the interval $\left(\text{−}\infty ,\infty \right)$ into the smaller intervals $\left(\text{−}\infty ,\mathrm{-1}\right),\left(\mathrm{-1},0\right),\left(0,1\right),$ and $\left(1,\infty \right).$

Step 2: Since $f^{\prime }$ is continuous over each subinterval, it suffices to choose a test point $x$ in each of the intervals from step $1$ and determine the sign of $f^{\prime }$ at each of these points. The points $x=\mathrm{-2},x=-\frac{1}{2},x=\frac{1}{2},\text{and}x=2$ are test points for these intervals.

Interval	Test Point	Sign of $f^{\prime }\left(x\right)=\frac{5\left(1-x^{4\text{/}3}\right)}{3x^{2\text{/}3}}$ at Test Point	Conclusion
$\left(\text{−}\infty ,\mathrm{-1}\right)$	$x=\mathrm{-2}$	$\frac{\left(\text{+}\right)\left(\text{−}\right)}{+}=\text{−}$	$f$ is decreasing.
$\left(\mathrm{-1},0\right)$	$x=-\frac{1}{2}$	$\frac{\left(\text{+}\right)\left(\text{+}\right)}{+}=+$	$f$ is increasing.
$\left(0,1\right)$	$x=\frac{1}{2}$	$\frac{\left(\text{+}\right)\left(\text{+}\right)}{+}=+$	$f$ is increasing.
$\left(1,\infty \right)$	$x=2$	$\frac{\left(\text{+}\right)\left(\text{−}\right)}{+}=\text{−}$	$f$ is decreasing.

Step 3: Since $f$ is decreasing over the interval $\left(\text{−}\infty ,\mathrm{-1}\right)$ and increasing over the interval $\left(\mathrm{-1},0\right),$ $f$ has a local minimum at $x=\mathrm{-1}.$ Since $f$ is increasing over the interval $\left(\mathrm{-1},0\right)$ and the interval $\left(0,1\right),$ $f$ does not have a local extremum at $x=0.$ Since $f$ is increasing over the interval $\left(0,1\right)$ and decreasing over the interval $\left(1,\infty \right),f$ has a local maximum at $x=1.$ The analytical results agree with the following graph.

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Concavity and points of inflection

We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity    of the function.

[link] (a) shows a function $f$ with a graph that curves upward. As $x$ increases, the slope of the tangent line increases. Thus, since the derivative increases as $x$ increases, $f^{\prime }$ is an increasing function. We say this function $f$ is concave up. [link] (b) shows a function $f$ that curves downward. As $x$ increases, the slope of the tangent line decreases. Since the derivative decreases as $x$ increases, $f^{\prime }$ is a decreasing function. We say this function $f$ is concave down.

In general, without having the graph of a function $f,$ how can we determine its concavity? By definition, a function $f$ is concave up if $f^{\prime }$ is increasing. From Corollary $3,$ we know that if $f^{\prime }$ is a differentiable function, then $f^{\prime }$ is increasing if its derivative $f\text{″}\left(x\right)>0.$ Therefore, a function $f$ that is twice differentiable is concave up when $f\text{″}\left(x\right)>0.$ Similarly, a function $f$ is concave down if $f^{\prime }$ is decreasing. We know that a differentiable function $f^{\prime }$ is decreasing if its derivative $f\text{″}\left(x\right)<0.$ Therefore, a twice-differentiable function $f$ is concave down when $f\text{″}\left(x\right)<0.$ Applying this logic is known as the concavity test    .