Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Concavity and Points of Inflection

A function may be concave up or down. In Figure 5.13, below, the function is concave up (abbreviated CU) on the interval ${\displaystyle \left[a,b\right]}$ and concave down (abbreviated CD) on the interval ${\displaystyle \left[b,c\right]}$.

Figure 5.13. Function showing concave up and concave down segments

The graphs in Figure 5.14, below, show two functions that are concave up and down on the interval ${\displaystyle \left[c,d\right]}$. Observe that these functions satisfy the conditions of Definition 5.11 for any $a$ and $b$ in ${\displaystyle \left[c,d\right]}$.

Figure 5.14. Functions that are concave up and concave down on the interval ${\displaystyle \left[c,d\right]}$

The graphs in Figure 5.15, below, show two functions: one is concave up on the interval ${\displaystyle \left(c,d\right)}$, and the other is concave down on the interval ${\displaystyle \left(-\infty ,d\right)}$. Observe that these functions satisfy the conditions of Definition 5.11 for any $a$ and $b$ in the corresponding intervals.

Figure 5.15. Functions that are concave up on the interval ${\displaystyle \left(c,d\right)}$ and concave down on the interval ${\displaystyle \left[-\infty ,d\right)}$

The graphs in Figure 5.16, below, show two functions: one is concave up on the interval ${\displaystyle \left[c,d\right)}$, and the other is concave down on the interval ${\displaystyle \left[c,\infty \right)}$. Observe that these functions satisfy the conditions of Definition 5.11 for any $a$ and $b$ in the corresponding intervals.

Figure 5.16. Functions that are concave up on the interval ${\displaystyle \left[c,d\right)}$ and concave down on the interval ${\displaystyle \left[c,\infty \right)}$

The graphs in Figure 5.17, below, show two functions: one is concave up on the interval ${\displaystyle \left(c,d\right]}$ and the other is concave down on the interval ${\displaystyle \left(-\infty ,d\right])}$. Observe that these functions satisfy the conditions of Definition 5.11 for any $a$ and $b$ in the corresponding intervals.

Figure 5.17. Functions that are concave up on the interval ${\displaystyle \left(c,d\right]}$ and concave down on the interval ${\displaystyle \left(-\infty ,d\right]}$

Compare Definition 5.11 of this unit, with the equivalent informal definition given on page 213 of the textbook. Note that a function $f$ is concave up (down) if and only if the graph of $f$ lies above (below) each tangent line (wherever they exist), except at the point of contact.

Example 5.38. The quadratic function ${\displaystyle f\left(x\right)=x^2}$ is concave up on its domain.

The cubic function ${\displaystyle g\left(x\right)=x^3}$ is concave down on ${\displaystyle \left(-\infty ,0\right]}$, and concave up on the interval ${\displaystyle \left[0,\infty \right)}$.

Example 5.39. The tangent function is concave down on the interval ${\displaystyle \left(-{\displaystyle \frac{\pi }{2}},0\right]}$, and concave up on the interval ${\displaystyle \left[0,{\displaystyle \frac{\pi }{2}}\right)}$

To find the intervals of concavity—that is, where the function is concave up or down—we must observe that if the function is concave up, then the tangent lines are increasing.

Figure 5.18. Function that is concave up; tangent lines have negative slope

In Figure 5.18, above, the slopes of the tangent lines ${\displaystyle S_1,S_2}$ and ${\displaystyle S_3}$ are negative, ${\displaystyle S_4=0}$, and ${\displaystyle S_1<S_2<S_3<S_4=0}$.

Figure 5.19. Function that is concave up; tangent lines have positive slope

In Figure 5.19, above, the slopes of the tangent lines are positive, ${\displaystyle S_4=0}$, and ${\displaystyle 0=S_4<S_5<S_6<S_7<S_8}$.

In both figures, the slopes are increasing.

Therefore, when the function $f$ is concave up, its derivative function ${\displaystyle f^{\prime }}$ is increasing, and the derivative of ${\displaystyle f^{\prime }}$—that is the second derivative function, ${\displaystyle f^{″}}$—is positive.

If the function is concave down, then the tangent lines are decreasing.

Figure 5.20. Function that is concave down; tangent lines have positive slope

In Figure 5.20, above, the slopes of the tangent lines ${\displaystyle S_1}$ and ${\displaystyle S_2}$ are positive, ${\displaystyle S_3=0}$ and ${\displaystyle S_1>S_2>S_3=0}$.

Figure 5.21. Function that is concave down; tangent lines have negative slope

In Figure 5.21, above, the slopes of the tangent lines ${\displaystyle S_4}$ and ${\displaystyle S_5}$ are negative, ${\displaystyle S_3=0}$ and ${\displaystyle 0=S_3>S_4>S_5}$.

In both figures the slopes are decreasing.

When the function $f$ is concave down, its derivative function ${\displaystyle f^{\prime }}$ is decreasing, and the derivative of ${\displaystyle f^{\prime }}$—that is, the second derivative function, ${\displaystyle f^{″}}$—is negative.

These observations result in the following concavity test.

Clearly, therefore, to find the concavity of a function $f$, we must find the intervals where the second derivative is positive or negative. To do so, we

1. find the critical numbers of the derivative function (i.e., all $c$ in the domain of ${\displaystyle f^{\prime }}$, such that either ${\displaystyle f^{″}\left(c\right)=0}$ or ${\displaystyle f^{″}\left(c\right)}$ does not exist); then,
2. simplify and factor the second derivative function as much as possible so that we can
3. make up a table with the intervals given by the critical numbers and the factors that determine the sign of the second derivative function.

Example 5.40. In Example 5.24, we found that the derivative of

${\displaystyle f\left(x\right)=\sqrt{x^2-9}}$

is

${\displaystyle f^{\prime }\left(x\right)={\displaystyle \frac{x}{\sqrt{x^2-9}}}\text{.}}$

Hence,

${\displaystyle f^{″}(x)={\displaystyle \frac{\sqrt{x^2-9}-x \left({\displaystyle {\displaystyle \frac{x}{\sqrt{x^2-9}}}}\right)}{x^2-9}}=-{\displaystyle \frac{9}{\sqrt{{\left(x^2-9\right)}^3}}}}$

We see that ${\displaystyle f^{″}\left(x\right)<0}$ for all $x$ in the domain of $f$, so the function is concave down in its domain.

Example 5.41. In Example 5.26, we found the derivative

${\displaystyle y^{\prime }=3x^2-12x-15=3\left(x^2-4x-5\right)\text{.}}$

Hence,

${\displaystyle y^{″}\left(x\right)=6x-12=6\left(x-2\right)=0\text{ for }x=2\text{,}}$

${\displaystyle y^{″}\left(x\right)>0\text{ for }x>2}$

and

${\displaystyle y^{″}\left(x\right)<0\text{ for }x<2\text{.}}$

So the function is concave up on the interval ${\displaystyle \left(2,\infty \right)}$, and concave down on the interval ${\displaystyle \left(-\infty ,2\right)}$

Example 5.42. Check that the second derivative function of

${\displaystyle g\left(x\right)=x\sqrt{x+2}}$

is

${\displaystyle g^{″}\left(x\right)={\displaystyle \frac{3x+8}{4{\left(x+2\right)}^{3∕2}}}\text{.}}$

So, ${\displaystyle g^{″}\left(x\right)=0}$ if ${\displaystyle x=-{\displaystyle \frac{8}{3}}}$.

The domain of the function is the interval ${\displaystyle \left[-2,\infty \right)}$, and ${\displaystyle x=-{\displaystyle \frac{8}{3}}}$ is not on this interval. We have ${\displaystyle g^{″}\left(x\right)>0}$ if ${\displaystyle x\ge -2}$.

Hence, the function is concave up on the interval ${\displaystyle \left[-2,\infty \right)}$.

Example 5.43. Check that the second derivative of the function

${\displaystyle h\left(x\right)=x^6-3x^4+3x^2}$

is

${\displaystyle h^{″}\left(x\right)=6\left(5x^4-6x^2+1\right)\text{.}}$

We factor this function, using the quadratic formula, setting ${\displaystyle u=x^2}$, so that

${\displaystyle 5x^3-6x^2+1=5u^2-6u+1=(u-1)(5u-1)}$

and

$h^{″}(x)=6(x^2-1)(5x^2-1)=6(x+1)(x-1)(\sqrt{5}x+1)(\sqrt{5}x-1)\text{.}$

The critical points of ${\displaystyle h^{\prime }}$ are

${\displaystyle x=1\text{,}x=-1\text{,}x={\displaystyle \frac{1}{\sqrt{5}}}\text{and}x=-{\displaystyle \frac{1}{\sqrt{5}}}\text{.}}$

The table we must create to find the sign of ${\displaystyle h^{″}}$ is shown below.

Hence, the function is concave up on the intervals

${\displaystyle \left(-\infty ,-1\right)\text{,}\left({\displaystyle \frac{-1}{\sqrt{5}}},{\displaystyle \frac{1}{\sqrt{5}}}\right)\text{and}\left(1,\infty \right);}$

it is concave down on the intervals

${\displaystyle \left(-1,-{\displaystyle \frac{1}{\sqrt{5}}}\right)\text{and}\left({\displaystyle \frac{1}{\sqrt{5}}},1\right)\text{.}}$

Definition 5.13. A continuous function $f$ on an interval $J$ has an inflection point at ${\displaystyle \left(a,f\left(a\right)\right)}$ if the function is concave up (down) on some subinterval ${\displaystyle \left(c,a\right)}$ of $J$ and it is concave down (up) on some subinterval ${\displaystyle \left(a,b\right)}$ of $J$.

Example 5.44. By Example 5.40, the function in Example 5.24 has no inflection points.

Example 5.45. By Example 5.41, the function in Example 5.26 has an inflection point at ${\displaystyle \left(2,y\left(2\right)\right)=\left(2,-42\right)}$

Example 5.46. By Example 5.42, the function of ${\displaystyle g\left(x\right)=x\sqrt{x+2}}$ has no inflection points.

Observe that at an inflection point is precisely the instant at which the derivative of the function changes from increasing (decreasing) to decreasing (increasing). This fact has two important consequences:

1. at an inflection point, the rate of change changes from increasing (decreasing) to decreasing (increasing).
2. at an inflection point, the second derivative function has a local extreme point.

Example 5.47. If the function ${\displaystyle y=x^3-6x^2-15x+4}$ in Example 5.26 is the position function of an object, then, by Example 5.41, at time ${\displaystyle x=2}$, the velocity of the object changed from increasing to decreasing. The acceleration is at its maximum at this instant.

Geometrically the second derivative of this function has a local maximum at ${\displaystyle x=2}$ (see Example 5.41).