The First Derivative Test

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We use the first derivative to help locate extrema.

The First Derivative Test

The method of the previous section for deciding whether there is a local maximum or minimum at a critical point by testing ‘‘near-by’’ points is not always convenient. Instead, since we have already had to compute the derivative to find the critical points, we can use information about the derivative to decide. Recall that

If $f'(x) >0$ on an interval, then $f$ is increasing on that interval.
If $f'(x) <0$ on an interval, then $f$ is decreasing on that interval.

So how exactly does the derivative tell us whether there is a maximum, minimum, or neither at a point? That’s where the first derivative test comes into play.

First Derivative Test Suppose that $f$ is continuous on an interval, and that $f'(a)=0$ or $f'(a)$ is undefined for some value of $a$ in that interval.

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is the location of a local maximum.
If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of $a$ , then $f(a)$ is the location of a local minimum.
If $f'(x)$ has the same sign to the left and right of $a$ , then $f(a)$ is not the location of a local extremum.

Consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Find the intervals on which $f$ is increasing and decreasing and identify the local extrema of $f$ .

Start by computing $\ddx f(x) = \answer [given]{x^3+x^2-2x}.$ Now we need to find when this function is positive and when it is negative. To do this, solve $f'(x) = \answer [given]{x^3+x^2-2x} =0.$ Factor $f'(x)$ $\begin{align*} f'(x) &= \answer[given]{x^3+x^2-2x} \\ &=x(\answer[given]{x^2+x-2})\\ &=x(x+2)\answer[given]{(x-1)}. \end{align*}$ So the critical points (when $f'(x)=0$ or is undefined) are when $x=-2$ , $x=0$ , and $x=1$ . Now we can check points between the critical points to find when $f'(x)$ is increasing and decreasing: $\begin{align*} f'(-3)&=\answer[given]{-12},\\ f'(.5)&=\answer[given]{-0.625},\\ f'(-1)&=\answer[given]{2},\\ f'(2)&=\answer[given]{8}. \end{align*}$ From this we can make a sign table:

Hence $f$ is increasing on $(-2,0)$ and $(1,\infty )$ and $f$ is decreasing on $(-\infty ,-2)$ and $(0,1)$ . Moreover, from the first derivative test, the local maximum is at $x=0$ while the local minima are at $x=-2$ and $x=1$ , see the graphs of of $f(x) =x^4/4 + x^3/3 -x^2$ and $f'(x) = x^3 + x^2 -2x$ .

If you look carefully at the previous example, you’ll see that if $f'$ is zero and $f'$ is increasing at a point, then $f$ has a local minimum at the point. Alternatively, if $f'$ is zero $f'$ is decreasing at a point, then $f$ has a local maximum at the point. Thus, we see that we can gain information about $f$ by studying how $f'$ changes, namely whether $f'$ is increasing or decreasing. This leads us to our next section about the concavity of $f$ .