Short and Sweet Calculus

## 4.3 First Derivative Test For Local Extrema

We showed that if $$f(c)$$ is a local maximum or local minimum, then $$x=c$$ is a critical number; that is, $$f'(c)=0$$ or $$f'(c)$$ does not exist. However, we did not discuss how to determine if $$f(c)$$ is a local maximum, a local minimum, or neither. In this section, we provide a method for that.

Before we start, recall that if $$f'(x)>0$$ on an interval, then $$f$$ is an increasing function on that interval, and if $$f'(x)<0$$, then $$f$$ is a decreasing function (see Section [sec:ch6-Increasing-and-decreasing]).

• As we move along the curve of a function from left to right, if the sign of the derivative changes from $$+$$ to $$-$$ because the function changes from increasing $$\nearrow$$ to decreasing $$\searrow$$ , the function has a local (or relative) maximum at the critical point (see Figure 4.5).

• As we move along the curve from left to right, if the sign of the derivative changes from $$-$$ to $$+$$ because the function changes from decreasing $$\searrow$$ to increasing $$\nearrow$$, the function has a local (or relative) minimum at the critical point (see Figure 4.6).

• As we move along the curve from left to right, if the sign of the derivative to the left and to the right of a critical point is the same, the function has no local maximum or minimum at the critical point (see Figure 4.7).

More precisely, we have the following theorem.

4.3. (First Derivative Test) Assume $$f$$ is continuous at the critical point $$c$$.

(a) If $$f’$$ is $$+$$ to the left of $$c$$ and is $$-$$ to the right of $$c$$, then $$f$$ has a local maximum at $$c$$.

(b) If $$f’$$ is $$-$$ to the left of $$c$$ and is $$+$$ to the right of $$c$$, then $$f$$ has local minimum at $$c$$.

(c) If the sign of $$f’$$ is the same to the right and left of $$c$$, then $$f$$ does NOT have a local extremum at $$c$$.

• To remember the First Derivative Test, visualize $$\nearrow\searrow$$ for the case where $$f’$$ varies from $$+$$ to $$-$$ and $$\searrow\nearrow$$ for the case where $$f’$$ varies from $$-$$ to $$+$$.

• Recall that the critical numbers are the only possible locations for local extrema.

Example 4.3. Find the local maxima and minima of $f(x)=\frac{1}{2}x^{3}-\frac{3}{2}x^{2}+1.$

Solution

Because $$f$$ is a polynomial, it is differentiable everywhere, so the critical points are the points where $$f'(x)=0$$. $f(x)=\frac{1}{2}x^{3}-\frac{3}{2}x^{2}+1$ $\Rightarrow f'(x)=\frac{3}{2}x^{2}-3x=\frac{3}{2}x(x-2).$ Critical numbers: $f'(x)=0\Longleftrightarrow x=0,x=2.$

$$x$$ $$-\infty$$ 0 2 $$+\infty$$
sign of $$x$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$
sign of $$x-2$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$
$$\therefore$$ sign of $$f'(x)$$ $$+$$ $$0$$ $$-$$ $$0$$ $$+$$
Increasing/Decreasing $$f(x)$$ $$\nearrow$$ max $$\searrow$$ min $$\nearrow$$

As this sign table shows, because the sign of $$f’$$ changes from $$+$$ to $$-$$ at $$x=0$$, the function has a local maximum there, and because the sign of $$f’$$ changes from $$-$$ to $$+$$ at $$x=2$$, the function has a local minimum at $$x=2$$. The graph of $$f$$ is shown in Figure 4.8.

Example 4.4. Investigate the local extrema of $f(x)=(x+1)^{2}(x-2)^{3}.$

Solution

$$f$$ is a differentiable function everywhere (because it is the product of two differentiable functions $$(x+1)^{2}$$ and $$(x-2)^{3}$$). Therefore, the only critical numbers of $$f$$ are the zeros of $$f’$$. $f(x)=\underbrace{(x+1)^{2}}_{u}\underbrace{(x-2)^{3}}_{v}$ \begin{aligned} \Rightarrow f'(x) & =\underbrace{2(x+1)}_{u’}\underbrace{(x-2)^{3}}_{v}+\underbrace{(x+1)^{2}}_{u}\underbrace{3(x-2)^{2}}_{v’}\\ & =(x+1)(x-2)\left[2(x-2)+3(x+1)\right]\tag{factor}\\ & =(x+1)(x-2)^{2}(5x-1)\end{aligned} Critical numbers: $f'(x)=0\Longleftrightarrow x=-1,x=2,x=\frac{1}{5}.$ Now we need to determine the sign of $$f'(x)$$:

$$x$$ $$-\infty$$ $$-1$$ $$\frac{1}{5}$$ $$2$$ $$+\infty$$
sign of $$x+1$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$ $$+$$ $$+$$
sign of $$(x-2)^{2}$$ $$+$$ $$+$$ $$+$$ $$+$$ $$+$$ $$0$$ $$+$$
sign of $$5x-1$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$
$$\therefore$$ sign of $$f'(x)$$ $$+$$ $$0$$ $$-$$ $$0$$ $$+$$ $$0$$ $$+$$
Increasing/Decreasing $$f(x)$$ $$\nearrow$$ max $$\searrow$$ min $$\nearrow$$ $$\nearrow$$

This sign table shows that $$f$$ has a local maximum at $$x=-1$$ and a local minimum at $$x=-5/4$$. Because the sign of $$f’$$ does not change in moving from left to right through $$x=1$$, $$f$$ does not have a local extremum there. In fact, $$(2,f(2))$$ is a point of inflection, as we can see from the graph of $$f$$ (Figure 4.9).

To determine the sign of the derivative at points near a particular critical point, an alternative method to the sign table, is to substitute in the formula of the derivative, first, a value of the variable just a little less than the corresponding critical value, and then one a little greater.