Calculus I 03.04 Concavity and the Second Derivative Test

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3.4 Concavity and the Second Derivative Test

• Determine intervals on which a function is concave upward or concave downward.
• Find any inflection points on the graph for a function.
• Apply the Second Derivative Test to find relative extrema for a function.

Concavity

Locating the intervals where \({f}'\) increases or decreases can be used to determine where the graph for \(f\) is curving upward or curving downward.

Concavity for \(f\) is related to how the derivative slopes. Figure 3.4.1

Definition 3.4.1 Concavity Let \(f\) be differentiable on an open interval \(I\). The graph for \(f\) is concave upward on \(I\) when \({f}'\) is increasing on the interval and concave downward on \(I\) when \({f}'\) is decreasing on the interval. 1. Let \(f\) be differentiable on an open interval \(I\). If the graph for \(f\) is concave upward on \(I\), then the graph for \(f\) lies above all its tangent lines on \(I\), as shown in Figure 3.4.2(a). 2. Let \(f\) be differentiable on an open interval \(I\). If the graph for \(f\) is concave downward on \(I\), then the graph for \(f\) lies below all its tangent lines on \(I\), as shown in Figure 3.4.2(b). Figure 3.4.2 To find the open intervals on which the graph for a function \(f\) is concave upward or concave downward, first find the intervals on which \({f}'\) is increasing or decreasing. For example, the graph for $$f(x)=\frac{1}{3}x^3-x$$ is concave downward on the open interval \((-\infty,0)\) because \({f}'(x)=x^2-1\) is decreasing there, as shown in Figure 3.4.1. Similarly, the graph for \(f\) is concave upward on the interval \((0,\infty)\) because \({f}'\) is increasing there.

Theorem 3.4.1 shows how to use the second derivative for a function \(f\) to determine intervals on which the graph for \(f\) is concave upward or downward. The proof follows directly from Theorem 3.3.1 and the concavity definition.

Theorem 3.4.1 Test for Concavity

Let \(f\) be a function whose second derivative exists on an open interval \(I\).
1. If \({f}(x)>0 for all \(x\) in \(I\), then the graph for \(f\) is concave upward on \(I\).
2. If \({f}(x)<0 for all \(x\) in \(I\), then the graph for \(f\) is concave downward on \(I\).
3. If \({f}(x)=0 for all \(x\) in \(I\), then the graph for \(f\) is not concave on \(I\).
To apply Theorem 3.4.1, locate the \(x\)-values at which \({f}''(x)=0\) or \({f}''\) does not exist. Use these \(x\)-values to determine the test intervals. Finally, test the sign for \({f}''(x)\) in each test intervals.

Example 3.4.1 Determining Concavity

The sign for \({f}''\) determines the concavity for \(f\)'s graph. Figure 3.4.3

Determine the open intervals on which the graph for $$f(x)=\frac{6}{x^2+3}$$ is concave upward or downward. Solution Begin by observing that \(f\) is continuous on the entire real number line. Next, find the second derivative for \(f\). \(f(x)\) \(=6(x^2+3)^{-1}\)     Rewrite original function \({f}'(x)\) \(=(-6)(x^2+3)^{-2}(2x)\)     Differentiate $$=\frac{-12x}{(x^2+3)^2}$$     First derivative $${f}''(x)$$ $$=\frac{(x^2+3)^2(-12)-(-12x)(2)(x^2+3)(2x)}{(x^2+3)^4}$$     Differentiate $$=\frac{36(x^2-1)}{(x^2+3)^3}$$     Second Derivative Because \({f}''=0\) when \(x= \pm 1\) and \({f}''\) is defined on the entire real number line. Test \({f}''\) on the intervals \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). The results are shown in Table 3.4.1 and in Figure 3.4.3. Table 3.4.1 Interval \(-\infty<x<-1\) \(-1<x<1\) \(1<x<\infty\) Test Value \(x=-2\) \(x=0\) \(x=2\) Sign for \({f}''(x)\) \({f}''(-2)>0\) \({f}''(0)<0\) \({f}''(2)>0\) Conclusion Concave Upward Concave Downward Concave Upward

The function given in Example 3.4.1 is continuous on the entire real number line. When there are \(x\)-values at which the function is not continuous, these values should be used, along with the points at which \({f}''(x)=0\) or \({f}''(x)\) does not exist, to form the test intervals

Example 3.4.2 Determining Concavity

Figure 3.4.4

Determine the open intervals on which the graph for $$f(x)=\frac{x^2+1}{x^2-4}$$ is concave upward or downward. Solution Differentiating twice produces the following. \(f(x)\) $$=\frac{x^2+1}{x^2-4}$$     Rewrite original function \({f}'(x)\) $$=\frac{(x^2-4)(2x)-(x^2+1)(2x)}{(x^2-4)^4}$$     Differentiate $$=\frac{-10x}{(x^2-4)^2}$$     First derivative $${f}''(x)$$ $$=\frac{(x^2-4)^2(-10)-(-10x)(2)(x^2-4)(2x)}{(x^2-4)^4}$$     Differentiate $$=\frac{10(3x^2+4)}{(x^2-4)^3}$$     Second Derivative There are no points at which \({f}''(x)=0\), but at \(x=\pm 2\), the function \(f\) is not continuous. Test for concavity in the intervals \((-\infty,-2)\), \((-2,2)\), and \((2,\infty)\) as shown in Table 3.4.2. The graph for \(f\) is shown in Figure 3.4.4. Table 3.4.2 Interval \(-\infty<x<-2\) \(-12<x<2\) \(2<x<\infty\) Test Value \(x=-3) \(x=0\) \(x=3\) Sign for \({f}''(x)\) \({f}''(x)>-3\) \({f}''(0)<0\) \({f}''(3)>0\) Conclusion Concave Upward Concave Downward Concave Upward

Inflection Point

Concavity for \(f\) changes at an inflection point. Note the graph crosses its tangent line at an inflection point. Figure 3.4.5 Inflection points can occur when \({f}''(x)=0\) or \({f}''\) does not exist. Figure 3.4.6

The graph in Figure 3.4.3 has two points where the concavity changes. If the tangent line to the graph exists at such a point, then that point is an inflection point. Figure 3.4.5 shown inflection points in three types. Definition 3.4.2 Inflection Point Let \(f\) be a continuous function on an open interval. Let \(c\) be a point in that interval. If the graph for \(f\) has a tangent line at this point \((c,f(c))\), then this point is an inflection point. This point is where the concavity changes from upward to downward or the reverse. Some sources use a definition that does not require a tangent line at the inflection point. For example, the function $$f(x)=\left\{\begin{matrix} x^3, & x<0\\ x^2+2x, & x \geq 0 \end{matrix}\right. $$ does not have an inflection point at the origin, even though the concavity for the graph changes from downward to upward. To locate possible inflection points determine the \(x\)-values for which \({f}''(x)=0 \text{ or } {f}''(x)\) does not exist. This is similar to locating extrema for \(f\). Theorem 3.4.2. Inflection Point If \((c,f(c))\) is and inflection point on the graph for \(f\), then either \({f}''(c)=0\) or \({f}''\) does not exist at \(x=c\). Example 3.4.3 Finding Inflection Points Determine the inflection points and discuss the concavity for the graph for \(f(x)=x^4-4x^3.\) Solution Differentiating twice produces the following. \(f(x)\) \(=x^4-4x^3\) \({f}'(x)\) \(=4x^3-12x^2\)     First derivative \({f}''(x)\) \(=12x^2-24x=12x(x-2)\)     Second derivative Solving \({f}''(x)=0\) produces possible inflection points at \(x=0\) and \(x=2\). Testing the intervals between these \(x\)-values determines they both yield inflection points.The testing is shown below in Table 3.4.3. The graph for \(f\) is shown in Figure 3.4.6. Table 3.4.3 Interval \(-\infty<x<0\) \(0<x<2\) \(2<x<\infty\) Test Value \(x=-1) \(x=1\) \(x=3\) Sign for \({f}''(x)\) \({f}''(-1)>0\) \({f}''(1)<0\) \({f}''(3)>0\) Conclusion Concave Upward Concave Downward Concave Upward

\({f}''(x)=0\), but \((0,0)\) is not an inflection point. Figure 3.4.7

The converse for Theorem 3.4.2 is not generally true. It is possible for the second derivative to equal zero at a point that is not an inflection point. For example, the graph for \(f(x)=x^4\) is shown in Figure 3.4.7. The second derivative is zero when \(x=0\), but the point \((0,0)\) is not an inflection point because the graph is concave upward in both intervals \(-\infty < x < 0\) and \(0<x<\infty\).

The Second Derivative Test

Figure 3.4.8

The second derivative is used to test for concavity and for relative extrema. If the graph for a differentiable function \(f\) is concave upward on an open interval containing \(c\), and \({f}'(c)=0\), then \(f(c)\) must be a relative minimum for \(f\). Conversely, if the graph for a differentiable function \(f\) is concave downward on an open interval containing \(c\), and \({f}'(c)=0\), then \(f(c)\) must be a relative maximum for \(f\),as shown in Figure 3.4.8. This is defined in Theorem 3.4.3. Theorem 3.4.3 Second Derivative Test Let \(f\) be a differential function such that \({f}'(c)=0\) and the second derivative for \(f\) exists on an open interval containing \(c\). 1. If \({f}''(x)>0\), then \(f\) has a relative minimum at \((c,f(c))\). 2. If \({f}''(x)<0\), then \(f\) has a relative maximum at \((c,f(c))\). If \({f}''(x)=0\) the test fails. Then \(f\) may have a relative maximum, a relative minimum, or neither. In this case use the First Derivative Test. Proof Proof for Case 1. If \({f}'(c)=0\) and \({f}''(c)>0\), then there exists an open interval \(I\) containing \(c\) for which $$\frac{{f}'(x)-{f}'(c)}{x-c}=\frac{{f}'(x)}{x-c}>0$$ for all \(x \neq c\) in \(I\). If \(x<c\), then \(x-c<0\) and \({f}'(x)<0\). If \(x>c\), then \(x-c>0\) and \({f}'(x)>0\). Therefore, \({f}'(x)\) changes from negative to positive at \(c\), and the First Derivative Test implies that \(f(c)\) is a relative minimum. The second case is proven in the same manner.

Example 3.4.4 Using the Second Derivative Test

The point \((0,0)\) is neither a relative minimum nor a relative maximum. Figure 3.4.9

Find the relative extrema for \(f(x)=-3x^5+5x^3.\) Solution Begin by finding the first derivative for \(f\). \({f}'(x)= -15x^4+15x^2=15x^2(1-x^2)\) The first derivative yields \(x=-1,\:0, \text{ and }1\) are the only critical numbers for \(f\). By finding the second derivative \({f}''(x)=-60x^3+30x=30x(1-2x^2)\) the Second Derivative Test can be applied as shown in Table 3.4.4 below. Table 3.4.4 Point \((-1,-2)\) \((0,0)\) \((1,2)\) Sign for \({f}''(x)=0\) \({f}''(-1)>0\) \({f}''(0)=0\) \({f}''(1)<0\) Conclusion Relative minimum Test fails Relative Maximum Because the Second Derivative Test fails at \((0,0)\), the First Derivative Test yields that \(f\) increases to the left and right fpr \(x=0\). Therefore, \((0,0)\) is neither a relative minimum nor a relative maximum even though the graph has a horizontal tangent line at this point. The graph for \(f\) is shown in Figure 3.4.9.

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Parent Article: Calculus I 03 Differentiation Applications