7.4 Concavity and Second Derivative Test

7.4.1 Concavity and Inflection Points

In 7.2 we studied the intervals where \(f’\) is positive, and those where it is negative. At these intervals, \(f\) is respectively, increasing, decreasing. And, called the points where \(f’\) is zero or undefined critical points. In this section we study the intervals where \(f”\) is positive, and those where it is negative. And, analyze what happens at points where \(f”\) is zero or undefined.

Definition 5. Let \(f\) be a differentiable function on an open interval \(I\), and let \(c\in I\).

1. \(f\) is said to be concave upward at \(\pmb{c}\) if there exists an open interval \((a,b)\subset I\) such that \(c\in (a,b)\), and the graph of \(f\) lies above the graph of the tangent line \(y = f'(x)(x-c) + f(c)\), that is, if
   \begin{equation}\label{ch07-04-eq01}
   f(x) \geq f'(x)(x-c) + f(c)\;\; \forall \; x\in (a,b).
   \end{equation}
2. \(f\) is said to be concave downward at \(\pmb{c}\) if there exists an open interval \((a,b)\subset I\) such that \(c\in (a,b)\), and the graph of \(f\) lies below the graph of the tangent line \(y = f'(x)(x-c) + f(c)\), that is, if
   \begin{equation}\label{ch07-04-eq02}
   f(x) \leq f'(x)(x-c) + f(c)\;\; \forall \; x\in (a,b).
   \end{equation}

Figure 7.19: Concavity and Inflection Point

Definition 6. Let \(f\) be a differentiable function on an open interval \(I\).

1. \(f\) is said to be concave upward on \(\pmb{I}\) if it is concave upward at each point \(c\in I\).
2. \(f\) is said to be concave downward on \(\pmb{I}\) if it is concave downward at each point \(c\in I\).

Figure 7.19 shows a the graph of a function that is concave upward for \(x < k\) and concave downward for \(x > k\).

Definition 7. Let \(f\) be a differentiable function on an open interval \(I\), and let \(c\in I\). \(c\) is called a point of inflection of \(\pmb{f}\), if there exists an interval \((a,b)\subset I\) with \(c\in (a,b)\) such that \(f\) changes concavity on the intervals \((a,c)\) and \((c,b)\), that is, \(f\) concave upward on \((a,c)\), and concave downward on \((c,b)\), or vice versa.

In Figure 7.19, the point \(k\) is a point of inflection.

Theorem 9. Let \(f\) be a function on an open interval \(I\), and assume that \(f”(x)\) exists for all \(x\in I\).

1. If \(f”(x) > 0\) for all \(x\in I\), then \(f\) is concave upward on \(I\).
2. If \(f”(x) < 0\) for all \(x\in I\), then \(f\) is concave downward on \(I\).

Proof (a) Let \(c\in I\), and define
\[L_c(x)=f(c) +f'(c) (x-c).\]
We need to show that there exists an open interval \((a,b)\subset I\) such that \(c\in (a,b)\) and that (7.2) is valid, that is,
\[f(x) \geq L_a(x), \;\; \forall \; x\in (a,b).\]
Alternatively, we need to show that
\[f(x) – L_a(x) \geq 0, \;\; \forall \; x\in (a,b).\]
Let
\[g(x) = f(x) – L_a(x), \;\; \forall \; x\in I.\]
Then, we have (explain why)
\[g(c) =0, \;\; g'(c) = 0, \;\; \text{ and }\;\; g”(x) = f”(x) > 0 \;\;\forall \; x\in I.\]
Thus, by Proposition~\ref{ch07-02-prop04}(a) \(g(0) = 0 \) is an absolute minimum value of \(g\) on \(I\). That is
\[g(x) \geq 0, \;\; \forall \; x\in I,\]
that is,
\[f(x) \geq L_a(x), \;\; \forall \; x\in I.\]
(b) The proof is similar and is left as an exercise (see #ch07-04-02-q10 in ch07-04-02).

Example 1.(Continuation of Example 1 in 7.2.1 and 7.2.3)
Given the function \(\displaystyle f(x) = \frac{x}{(x-2)^2}\),

(a) find the intervals where \(f\) is concave upward, and those intervals where concave downward;

(b) find its inflections points (if any).

Solution. Since
\[f”(x) = \frac{2(x+4)}{(x-2)^4},\]
the signs of \(f”\) are as in Figure~\ref{ch07-04-f02}(a) . Thus, \(f\) is concave upward on \((-4, 2)\cup ( 2, \infty)\), and concave downward on \((-\infty, -4)\). Furthermore, \(f\) has an inflection point at \(x=-4\), because it changes concavity at this point (see Figure~\ref{ch07-04-f02}(b)). The graph of \(f\) in Figure~\ref{ch07-04-f02}(c) shows the absolute minimum of \(f\) and the inflection point.

Figure 7.20: Example 1: (a)

Figure 7.20: Example 1: (b)

Figure 7.20: Example 1: (c)

Example 2. Given the function \(\displaystyle f(x) = x e^{-2 x^2}\), (a) find the intervals where \(f\) is concave upward, and those intervals where concave downward; (b) find its inflections points.

Solution. The first and second derivative of \(f\) are given by (verify this)

\[f'(x) = e^{-2 x^2}(1-4 x^2)\;\;\; \text{ and } \;\;\; f”(x) = e^{-2 x^2}x (4 x^2 – 3).\]

Since \(e^{-2 x^2} > 0\) for all \(x\in \mathbb{R}\), the signs of \(f”\) is determined by the signs of the polynomial \(x (4 x^2 – 3) = x (2 x – \sqrt{3})(2 x + \sqrt{3})\). Thus, the signs of \(f”\) are as in Figure 7.21(a). Thus,

\(f\) is concave upward on \(\displaystyle \left (-\frac{\sqrt {3}}{2}, 0\right )\cup \left ( \frac{\sqrt {3}}{2}, \infty\right )\), and downward on \(\displaystyle \left (-\infty, -\frac{\sqrt {3}}{2}\right )\cup \left (0, \frac{\sqrt {3}}{2}\right )\).

Furthermore, \(f\) has an inflection pointx at \(x=\pm \frac{\sqrt{3}}{2}\) and \(x=0\), because it changes concavity at this point (see Figure~\ref{ch07-04-f03}(a)), and the graph of \(f\) in Figure~\ref{ch07-04-f03}(b).

Figure 7.21: Example 2: (a)

Figure 7.21: Example 2: (b)

Example 3. (Continuation of Example 3 in 7.2.1 and 7.2.3.)
Given \(\displaystyle f(x) = \frac{1}{2}\sin 2x + \sin x\),
(a) find the intervals where \(f\) is concave upward, and those intervals where concave downward;
(b) find its inflections points.

Solution. (a) Since \(f”(x) = – \sin x – 4\sin x \cos x = -\sin x (1 + 4\cos x)\), thus

\[\hspace{.4in}\hphantom{(*)}
f”(x) = 0 \Longleftrightarrow \left [\renewcommand{\arraystretch}{2}\begin{array}{l}
\sin x = 0 \\ \displaystyle \cos x = -\frac{1}{4}
\end{array}\right .
\Longleftrightarrow
\left [\renewcommand{\arraystretch}{2}\begin{array}{l}
x= \pi k, \; k \in \mathbb{Z} \\ \displaystyle x = \pm\arccos\left (-\frac{1}{4} \right )+ 2\pi k, \; k\in \mathbb{Z}
\end{array}\right .. \hspace{.4in}(*)
\]

Use the Unit Circle to see how these solutions are obtained.

A simple way to determine where \(f”\) is positive or negative consists on drawing the graphs of the two factors \(y = -\sin x\) and \(y = 1+4\cos x\). This will allow us to see where their product is positive or negative. Then, we have

• – on the interval \(\displaystyle \left (0, \cos^{-1}\left (-\frac{14}{4}\right )\right )\), one factor is positive and one negative, thus \(f”\) is negative on this interval;
• – on the interval \(\displaystyle \left ( \cos^{-1}\left (-\frac{14}{4}\right ), \pi\right )\), both factors are negative, thus \(f”\) is positive on this interval;
• – on the interval \(\displaystyle \left (\pi, 2\pi – \cos^{-1}\left (-\frac{14}{4}\right )\right )\), one factor is positive and one negative, thus \(f”\) is negative on this interval;
• – on the interval \(\displaystyle \left ( 2\pi – \cos^{-1}\left (-\frac{14}{4}\right ), 2\pi\right )\), both factors are positive, thus \(f”\) is positive on this interval

Thus, the signs of \(f”\) are as in Figure ~\ref{ch07-04-f04}(b), where we have set \(\displaystyle x_1= \cos^{-1}\left (-\frac{14}{4}\right )\) and \(\displaystyle x_2= 2\pi – \cos^{-1}\left (-\frac{14}{4}\right )\).

Figure 7.22: Example 3: (a)

Figure 7.22: Example 3: (b)

(b) Since the the signs of \(f”\) alternate on the intervals where it does not vanish, it follows that the graph changes concavity at the points where \(f”(x) = 0\). Namely, at all points of the form \((x, f(x))\) with \(x\) as in \((*)\). These are the inflection points (see Figure  7.23).

Figure 7.23: Example 3

7.4.2 Exercises

1. Each graph represents the graph of \(f”\) for some function \(f\). (i) Find the inflection points of \(f\); \;(ii) determine the intervals where \(f\) is concave upward and the intervals where concave downward.

Figure 7.24: # 1(a)

Figure 7.24: # 1(b)

2. \(f(x) = x^3 – 3 x\).
3. \(\displaystyle f(x) = \frac{x^2}{x-2}\).
4. \(f(x) = x \ln x^2\).
5. \(f(x) = \cos^2 x + \cos 2x\), \(x\in \left [0,2\pi\right ]\).
6. \(\displaystyle f(x) = \frac{ \sqrt{x}}{3-2x}\).
7. \(f(x) = \arctan \left (x + 1\right )\).
8. \(\displaystyle f(x) =\int_1^x t^{t^2}\,dt\).
9. \(\displaystyle f(x) =\int_{x^2}^0 2^{-t^2} t \,dt\).
10. \(\displaystyle f(x) = \frac{x-2}{(x-1)^2}\).

11. 11. Find the inflection points of each \(f\). (a) \(f(x) = \arctan (x^2-4)\); (b) \(\displaystyle f(x) = \int_1^{x^2} x^{x-1}\,dx\); (c) \(f(x) = 2^{-x^2}x\).
    12. Find the values of \(a\), \(b\), \(c\) such that the function \(f(x)= 2x^3+ax^2+bx + c\) has an inflection point at \(x=-1\), a critical point at \(x=1\) and \(f(1) = 3\).
    13. Prove Theorem 2(b).

7.4.3 Second Derivative Test

Proposition 5(Second Derivative Test.)
Let \(f\) be a differentiable function defined on an open interval that contains \(c\) and is such that \(f”(c)\) exists.

1. If \(f'(c) = 0\) and \(f”(c) > 0\), then \(f(c) \) is a local minimum value of \(f\).
2. If \(f'(c) = 0\) and \(f”(c) < 0\), then \(f(c) \) is a local maximum value of \(f\).

Remark. Note that the case \(f'(c) = 0\) and \(f”(c) = 0\) is not in the proposition. In this case, no conclusion can be drawn. Namely, \(f(c) \) maybe (i) a local maximum value of \(f\), (ii) a local minimum value of \(f\), or (iii) neither. For example, \(f(x) = -x^4\) provides an example of (i) at \(c=0\), \(f(x) = x^4\) provides an example of (ii) at \(c=0\), and \(f(x) = x^3\) provides an example of (iii) at \(c=0\).

Proof
(a) Since \(f”(c) > 0\), we have

\[\lim_{x\to c} \frac{f'(x)-f'(c)}{x-c} > 0.\]

It follows that for all \(x\) sufficiently close to \(c\), say on the interval \(I = (c-\delta, c+\delta)\), for a small value of \(\delta >0\),

\[\frac{f'(x)-f'(c)}{x-c} > 0 \;\; \Longrightarrow \;\; f'(x)-f'(c) \; \text{ and } \; x-c \; \text{ have the same sign. } \; \]

Thus, if \(x\in I\) and \(x < c\), then \(x-c < 0 \), and \(f'(x)-f'(c) < 0 \). Hence \(f'(x) < f'(c) = 0\). It follows that \(f\) is decreasing for \(x \in I\), \(x < c\). Similarly, if \(x\in I\) and \(x > c\), then \(x-c > 0 \), and \(f'(x)-f'(c) > 0 \). Hence \(f'(x) > f'(c) = 0\). It follows that \(f\) is increasing for \(x \in I\), \(x > c\). Then, by the First Derivative Test (Proposition~\ref{ch07-02-prop02}), \(f(c)\) is a local minimum value of \(f\).(b) We leave the proof of (b) as an exercise (see 7.4.4 # 13).

Example 1. Use the second derivative to determine the location of all local extrema for the function \(\displaystyle f(x) = \frac{1}{8}(x-2)^2(x+1)^2\).

Solution.
Since \(f\) is a polynomial function, its local extrema occur at points where the derivative vanishes. Hence, we compute
\[f'(x) = \frac{1}{8}\left ( 2(x-2)(x+1)^2+ (x-2)^2\cdot 2(x+1)\right ) = \frac{1}{4}(x – 2) (x + 1) (2 x – 1).\]
Thus
\[f'(x) = 0 \;\; \Longleftrightarrow \;\; x\in \left \{-1, \frac{1}{2}, 2\right \}.\]
We now find (verify this)
\[ f”(x) = \frac{3}{4} (2 x^2-2 x-1)\;\; \text{ and } \;\; f”(-1) = \frac{9}{4}, \;\; f”\left (\frac{1}{2}\right )= – \frac{9}{8},\;\; f(2) = \frac{9}{4}.\]
It follows, from the Second Derivative Test, that \(f(-1) = 0\) is a local minimum value, \(f”\left (\frac{1}{2}\right )= \frac{81}{128}\) is a local maximum value, and, \(f(-1) = f(2) = 0\) is a local minimum value (see Figure 7.25).

Figure 7.25: Example

What about absolute maxmimum and minimum values? Is there a way to find these values using the second derivative test? Recall Proposition 4 in 7.2.5, which we restate here. Note that in this case, the assumption is that \(f”\) exists for all \(x\in (a,b)\), versus the assumption that \(f”(c)\) exists at a single point \(c\) for the Second Derivative Test.

Proposition 6
Let \( f \) be a function defined on an open interval \( (a,b) \), such that \(f”(x)\) exists for all \(x\in (a,b)\), and let \(f'(c) = 0\) for some \(c\in (a,b)\). Then

1. 1.  \( (f(c) \) is an absolute minimum value of \( f \) if \( f”(x) > 0 \) for all \( x\in (a, b) \);
   2.  \( f(c) \) is an absolute maximum value of \( f \) if \( f”(x) < 0 \) for for all \(x\in (a, b) \).

(a) Since \(f”(x) > 0\), then \(f'(x)\) is an increasing function on \((a,b) \), thus, \(f'(x) < 0\) for \(a < x < c\) and \(f'(x) > 0\) for \(c < x < b\). The result follows immediately from Proposition 3(a). The proof of (b) is similar. We leave the proof of (b) as an exercise (see 7.4.4 # 14)

Example 2
Find the absolute maximum of absolute minimum value (if any) of the function \(\displaystyle f(x) = x^2 + \frac{1}{x}\), \(x\in (0, \infty)\).

Solution.

We compute
\(f'(x) = 2x-\frac{1}{x^2} = \frac{2 x^3-1}{x^2}\;\; \text{ and }\;\; f”(x) = 2 +\frac{3}{x^3}.\)
Thus
\(f'(x) = 0 \;\; \Longleftrightarrow \;\; 2 x^3-1 = 0 \;\; \Longleftrightarrow \;\; x=\frac{1}{\sqrt[3]{2}}.\)
Since
\( f^{\prime \prime}(x)>0 \forall x \in(0, \infty) \underset{\text { Prop.}}{\Longrightarrow}_{6} f\left(\frac{1}{\sqrt[3]{2}}\right)=\frac{3}{\sqrt[3]{2^{2}}} \)
is an absolute minimum value of \(f\) on \( (0, \infty) \).

7.4.4 Exercises

In problems 1-4, for the function \( f \),

•  find \(f’\) and \(f”\);
•  use the second derivative test to locate the local extrema of \(f\);
• find the intervals where \(f\) is increasing and the intervals where \(f\) is decreasing. Use this information to find the local extrema of the function. Confront your answer with (b).
•  Find the intervals where \(f\) is concave upward, the intervals where \(f\) is concave downward, and the inflection points;
•  produce a sketch of the graph of \(f\).

1. \(f(x) = (x+3)^3 (x+5)\).
2. \(f(x) = (x-5)^2 (x+1)^2\).
3. \(\displaystyle f(x) = \frac{ x+5}{(x+4)^2}\).
4. \(\displaystyle f(x) = \frac{(x+2)^2}{(x+3)^2}\).

In problems 5-8, find the absolute maximum value or absolute minimum value (if any) of the function.

5. \(\displaystyle f(x) = x^2 + \frac{4}{x^2}\), \(x\in (0, \infty)\).
6. \(\displaystyle f(x) = \frac{x^2 + 3}{x+1}\), \(x\in (- \infty, -1)\).
7. \(\displaystyle f(x) = \frac{x^2 + 3}{(x+1)^2}\), \(x\in ( -1, 5)\).
8. \(\displaystyle f(x) = \frac{x^3}{(x-1)^2}\), \(x\in ( – \infty,1)\).

In problems 9-12, (a) locate, using the second derivative test, the local maximum values and the local minimum values of the function (if any); (b) locate the \(x\)-coordinate of the inflec-tion points.

9. \(f(x) = x^2e^{-4x}\).
10. \(f(x) = (2x^2-3x)e^{-x}\).
11. \(f(x) = e^{-1/x^2}\).
12. \(f(x) = x\ln x^2\).
13. Prove Proposition 5(b).
14. Prove Proposition 6(b).
15. Find a polynomial function that has inflections points at \(x=-2\) and \(x=1\). Hint: how should \(f”\) look like?