Concavity and the 2nd Derivative Test

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Here we examine what the second derivative tells us about the geometry of functions.

Concavity

We know that the sign of the derivative tells us whether a function is increasing or decreasing at some point. Likewise, the sign of the second derivative $f''(x)$ tells us whether $f'(x)$ is increasing or decreasing at $x$ . We summarize the consequences of this seemingly simple idea in the table below:

If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. It is worth summarizing what we have seen already in a single theorem.

Test for Concavity Suppose that $f''(x)$ exists on an interval.

(a): $f''(x)>0$ on that interval whenever $y=f(x)$ is concave up on that interval.
(b): $f''(x)<0$ on that interval whenever $y=f(x)$ is concave down on that interval.

Let $f$ be a continuous function and suppose that:

$f'(x) > 0$ for $-1< x<1$ .
$f'(x) < 0$ for $-2< x<-1$ and $1<x<2$ .
$f''(x) > 0$ for $-2<x<0$ and $1<x< 2$ .
$f''(x) < 0$ for $0<x< 1$ .

Sketch a possible graph of $f$ .

Start by marking where the derivative changes sign and indicate intervals where $f$ is increasing and intervals $f$ is decreasing. The function $f$ has a negative derivative from $-2$ to $x=\answer [given]{-1}.$ This means that $f$ is increasingdecreasing on this interval. The function $f$ has a positive derivative from $x=\answer [given]{-1}$ to $x=\answer [given]{1}.$ This means that $f$ is increasingdecreasing on this interval. Finally, The function $f$ has a negative derivative from $x=\answer [given]{1}$ to $2$ . This means that $f$ is increasingdecreasing on this interval.

Now we should sketch the concavity: concave upconcave down when the second derivative is positive, concave upconcave down when the second derivative is negative.

Finally, we can sketch our curve:

Inflection points

If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. Of particular interest are points at which the concavity changes from up to down or down to up.

If the concavity of $f$ changes either from up to down or down to up at $x=a$ and $f$ has a tangent line at $x=a$ , then $x=a$ is an inflection point of $f.$

It is instructive to see some examples of inflection points:

It is also instructive to see some non-examples of inflection points:

We identify inflection points by first finding $x$ such that $f''(x)$ is zero or undefined and then checking to see whether $f''(x)$ does in fact go from positive to negative or negative to positive at these points.

Even if $f''(a) = 0$ , the point determined by $x=a$ might not be an inflection point. You also need to check that concavity of $f$ changes on either side of $x=a.$

Describe the concavity of $f(x)=x^3-x$ .

To start, compute the first and second derivative of $f(x)$ with respect to $x$ , $f'(x)=\answer [given]{3x^2-1}\qquad \text {and}\qquad f''(x)=\answer [given]{6x}.$ Since $f''(0)=0$ , there is potentially an inflection point at $x=0$ . Using test points, we note the concavity does change from down to up, hence $x=0$ is an inflection point of $f.$ The curve is concave down for all $x<0$ and concave up for all $x>0$ , see the graphs of $f(x) = x^3-x$ and $f''(x) = 6x$ .

Note that we need to compute and analyze the second derivative to understand concavity, which can help us to identify whether critical points correspond to maxima or minima. Sometimes, rather than using the first derivative test for extrema, the second derivative test can also help you to identify extrema.

The second derivative test

Recall the first derivative test:

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is 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 a local minimum.

If $f'$ changes from positive to negative, then $f'$ is decreasing. In this case, $f''$ might be negative, and if in fact $f''$ is negative then $f'$ is definitely decreasing, so there is a local maximum at the point in question. On the other hand, if $f'$ changes from negative to positive, then $f'$ is increasing. Again, this means that $f''$ might be positive, and if in fact $f''$ is positive then $f'$ is definitely increasing, so there is a local minimum at the point in question. We summarize this as the second derivative test.

Second Derivative Test Suppose that $f''(x)$ is continuous on an open interval containing $x=a$ and that $f'(a)=0$ for some value of $a$ in that interval.

If $f''(a) <0$ , then $f$ has a local maximum at $a$ .
If $f''(a) >0$ , then $f$ has a local minimum at $a$ .
If $f''(a) =0$ , then the test is inconclusive. In this case, $f$ may or may not have a local extremum at $x=a$ , and the first derivative test should be used to find out.

In certain situations, when the second derivative is easy to calculate, the second derivative test is often the easiest way to identify local extrema. However, if the second derivative is difficult to calculate, you may want to stick with the first derivative test. Also, if $f(x)$ has a critical point, $x=a$ at which $f'(a)$ is undefined, the second derivative test does not apply, which means you’ll want to use the first derivative test instead. To add to this, even if the second derivative is easy to calculate, if it turns out that $f''(a) = 0$ , then $f(x)$ is neither concave up nor concave down at $x=a$ , so no conclusions can be made using concavity/the second derivative about whether $x=a$ corresponds to a local maximum or minimum.

Once again, consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Use the second derivative test, to locate the local extrema of $f$ .

Start by computing $f'(x) = \answer [given]{x^3 + x^2 -2x} \qquad \text {and}\qquad f''(x) = \answer [given]{3x^2 + 2x-2}.$ Using the same technique as we used before, we find that $f'(-2) = \answer [given]{0},\qquad f'(0) = \answer [given]{0}, \qquad f'(1) = \answer [given]{0}.$ Now we’ll attempt to use the second derivative test, $f''(-2) = \answer [given]{6}, \qquad f''(0) =\answer [given]{ -2}, \qquad f''(1) = \answer [given]{3}.$ Hence we see that $f$ has a local minimum at $x=-2$ , a local maximum at $x=0$ , and a local minimum at $x=1$ by applying the second derivative test. See below for a plot of $f(x) =x^4/4 + x^3/3 -x^2$ and $f''(x) = 3x^2 + 2x -2$ :

If $f''(a)=0$ , what does the second derivative test tell us?

The function has a local extrema at $x=a$ . The function does not have a local extrema at $x=a$ . It gives no information on whether $x=a$ is a local extremum.