## 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.

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