# Inflection Points and Concavity

What is the difference between the functions: taking the square of a number or squaring a number. Both of these functions are increasing. But one of them is growing must faster than the other. Comparing the growth of functions is the idea behind concavity and is foundational in computer science.

The Second Derivative Test provides a technique for classifying local extrema using the sign of the second derivative at a second-order critical number. To appreciate the Second Derivative Test, we first understand the meaning of concavity (as opposed to convex, concave means curving inward). We first discuss the concavity test and then the first and second derivative tests for finding local extreme values. We illustrate these tests with several examples and exercises are given at the end.

## First and Second Order Critical Points

Definition. Suppose $c$ is in the domain of a function $f.$ We will call $c$ a first-order critical number of $f$ when $f'(c)=0$ or $f'(c)$ does not exist and a second-order critical number of $f$ when $f”(c)=0$ or $f”(c)$ does not exist.

## Concavity and Inflection Points

Definition. Suppose a function $f$ is differentiable on an open interval $I.$

(1) If $f’$ is increasing on $I,$ then the graph of $f$ is called concave upward on $I.$

(2) If $f’$ is decreasing on $I,$ then the graph of $f$ is called concave downward on $I.$

If the graph of $f$ lies above all of its tangents on an interval $I,$ it is concave upward on $I.$ If the graph of $f$ lies below all of its tangents, it is concave downward on $I.$

Definition. A point $P(c,f(c))$ on a curve is called an inflection point of the graph of $f$ provided $f$ has a tangent line at $c$ and the concavity of $f$ changes at $x=c.$

## Concavity Test

Theorem. (Concavity Test) Suppose a function $f$ is twice differentiable on an interval $I.$

(1) If $f”(x)>0$ for all $x$ in $I,$ then the graph of $f$ is concave upward on $I.$

(2) If $f”(x)<0$ for all $x$ in $I,$ then the graph of $f$ is concave downward on $I.$

Example. Determine where the curve $$y=x^4-4x^3$$ is concave upward, where it is concave downward, and where the points of inflection are. A sketch of the curve $y=x^4-4x^3.$

Solution. The first and second derivatives of the function $f(x)=x^4-4x^3$ are \begin{equation}f'(x)=4x^3-12x^2=4x^2(x-3)\end{equation} and \begin{equation} f”(x)=12x^2-24x=12x(x-2).\end{equation} Since $f”$ is defined for all real numbers and $f”(x)=0$ only when $x=0$ and $x=2$ the second-order critical numbers of $f$ are $x=0$ and $x=2.$ We summarize the Concavity Test in the following table: \begin{equation} \begin{array}{c|c|c|l} \text{Interval} & f & f” & \text{Conclusion} \\ \hline x<0 & & + & \text{concave up} \\ x=0 & 0 & 0 & \text{inflection point} \\ 02 & & + & \text{concave up} \end{array} \end{equation} Therefore, $f$ is concave up on $(-\infty ,0)\cup (2,+\infty )$ and concave down on $(0,2).$ The points $(0,0)$ and $(2,-16)$ are inflection points.

Example. Determine where the curve $$y=x^3-3x+1$$ is concave upward and where it is concave downward. Find all inflection points, local extrema, and sketch the curve. A sketch of the function $y=x^3-3x+1.$

Solution. The first and second derivatives of the function $f(x)=x^3-3x+1$ are \begin{equation}f'(x)=3x^2-3=3(x-1)(x+1)\end{equation} and $f”(x)=6x.$ Since $f’$ and $f”$ are polynomials they are both defined for all real numbers and since $f'(x)=0$ and $f”(x)=0$ only when $x=\pm 1$ and $x=0$, respectively, the first-order critical numbers are $x=\pm 1$ and the second-order critical number is $x=0.$ We summarize the First Derivative Test and the Concavity Test in the following table. \begin{equation} \begin{array}{c|c|c|c|l} \text{Interval} & f & f’ & f” & \text{Conclusion} \\ \hline x<-1 & & + & & \text{increasing} \\ x=-1 & 3 & 0 & & \text{relative maximium} \\ -11 & & + & & \text{increasing} \\ x<0 & & & – & \text{concave down} \\ x=0 & 1 & & 0 & \text{inflection point} \\ x>0 & & & + & \text{concave up} \end{array} \end{equation} Therefore, the function $f$ has a local maximum at $(3,0)$ and a local minimum at $(1,-1).$ The function $f$ is increasing on $(-\infty ,-1)\cup (1,+\infty )$ and decreasing on $(-1,1).$ The point $(0,1)$ is an inflection point because $f$ is concave up on the interval $(0,+\infty )$ and concave down on $(-\infty ,0).$

## Second Derivative Test

Theorem. (Second Derivative Test) Suppose $f”$ in continuous on an open interval that contains $c$ with $f'(c)=0.$

(1) If $f”(c)>0,$ then $f$ has a relative (local) minimum at $c.$

(2) If $f”(c)<0,$ then $f$ has a relative (local) maximum at $c.$

The Second Derivative Test is inconclusive when both $f'(c)=0$ and $f”(c)=0.$ For example, if $f(x)=x^3$ and $g(x)=x^4,$ both $f'(0)=g'(0)=0$ and $f”(0)=g”(0)=0.$ The point $x=0$ is a minimum for $g$ but is neither a maximum nor a minimum for $f.$ For some functions the Second Derivative Test might be a straightforward method for determining whether a point is a local extrema, however for some functions the First Derivative Test is a necessity.

Example. Use the Second Derivative Test to determine whether each first-order critical number of the function $$f(x)=3x^5-5x^3+2$$ corresponds to a relative maximum, a relative minimum, or neither. A sketch of the function $f(x)=3x^5-5x^3+2.$

Solution. Since $f'(x)=15 x^4-15 x^2=15 (x-1) x^2 (x+1)$ is defined for all real numbers and $f'(x)=0$ when $x=0$ and $x=\pm 1$ the first-order critical numbers are $x=0$ and $x=\pm 1$. Since \begin{equation}f”(x)=60 x^3-30 x=30 x \left(2 x^2-1\right)\end{equation} is continuous for all real numbers we can apply the Second Derivative Test. Since $f”(-1)<0$ the point $(-1,4)$ is a local maximum and since $f”(1)>0$ the point $(1,0)$ is a local minimum. Since $f”(0)=0$ the Second Derivative Test is inconclusive at $x=0.$

Example. Use the Second Derivative Test to determine whether each critical number of the function $$f(x)= x+4/x$$ corresponds to a relative maximum, a relative minimum, or neither. A sketch of the function $f(x)= x+4/x.$

Solution. The first derivative of the function $f$ is \begin{equation}f'(x)=1-\frac{4}{x^2}=\frac{x^2-4}{x^2}=\frac{(x-2)(x+2)}{x^2}.\end{equation} The first derivative is defined for all real numbers except $x=0$ and $f'(x)=0$ only when $x=\pm 2$ the first-order critical numbers of $f$ are $x= \pm 2.$ Note that, even though $f'(0)$ is undefined, so is $f(0)$ and so $0$ is not a critical number. The second derivative of the function $f$ is $f”(x)=\frac{4}{x^3}$ and since $f”$ is continuous for all real numbers except $x=0$ we can apply the Second Derivative Test. Since $f”(2)>0$ the point $(2,4)$ is a local minimum and since $f”(-2)<0$ the point $(-2,-4)$ is a local maximum.

Example. For the function $$f(x)=x^2e^{-3x}$$ find all first-order and second-order critical numbers. Apply the First Derivative Test, Concavity Test, and the Second Derivative Test. Sketch the graph of the function. A sketch of the function $f(x)=x^2e^{-3x}.$

Solution. The first and second derivatives of the function $f$ are \begin{equation}
f'(x)=2x e^{-3x}-3x^2 e^{-3x}=e^{-3x}\left(2x-3x^2\right)=x(2-3x)e^{-3x}
\end{equation} and \begin{equation} f”(x)=e^{-3x}\left(9x^2-12x+2\right).
\end{equation} Since both the first and second derivatives are continuous for all real numbers the first-order and second-order critical numbers will be found by solving $f'(x)=0$ and $f”(x)=0$, respectively. Solving $f'(x)=0$ we find the first-order critical numbers to be $x=0$ and $x=\frac{2}{3}.$ Solving $f”(x)=0$ we find the second-order critical numbers to be $x=\frac{2}{3}\pm \frac{\sqrt{2}}{3}.$
Since $f$ is continuous we can apply the First Derivative Test as follows:
\begin{equation} \begin{array}{c|c|c|l} \text{Interval} & f & f’ & \text{Conclusion} \\ \hline x<0 & & – & \text{decreasing} \\ x=0 & 0 & 0 & \text{relative minimum} \\ 0\frac{2}{3} & & – & \text{decreasing} \end{array} \end{equation} Since $f”$ is continuous we can apply the Second Derivative Test as follows: since $f”(0)>0$ the point $(0,0)$ is a local minimum and since $f”\left(\frac{2}{3}\right)<0$ the point $\left(\frac{2}{3},f\left(\frac{2}{3}\right)\right)$ is a local maximum. Applying the Concavity Test we have \begin{equation} \begin{array}{c|c|c|l} \text{Interval} & f & f” & \text{Conclusion} \\ \hline x<\frac{2}{3}-\frac{\sqrt{2}}{3} & & + & \text{concave up} \\ x=\frac{2}{3}-\frac{\sqrt{2}}{3} & \frac{1}{9} \left(6-4 \sqrt{2}\right) e^{-2+\sqrt{2}} & 0 & \text{inflection point} \\ \frac{2}{3}-\frac{\sqrt{2}}{3}\frac{2}{3}+\frac{\sqrt{2}}{3} & & + & \text{concave up} \end{array} \end{equation}

## Exercises on Inflection Points and Concavity

Exercise. For each of the following functions identify the inflection points and local maxima and local minima. Identify the intervals on which the function is concave up and concave down. Sketch the graph showing these specific features.

$(1) \quad f(x)=\frac{x^4}{4}-2x^2+4$

$(2) \quad f(x)=\frac{9}{14}x^{1/3}\left(x^2-7\right)$

$(3) \quad f(x)=\tan x-4x$ on $-\frac{\pi }{2}<x<\frac{\pi }{2}$

$(4) \quad f(x)=2 \cos x-\sqrt{2}x$ on $-\pi <x<\frac{3\pi }{2}$

$(5) \quad f(x)=x^2-4x+3$

$(6) \quad f(x)=x^4-2x^2$

$(7) \quad f(x)=x-\sin x$ on $0\leq x\leq 2\pi$

$(8) \quad f(x)=\sqrt{|x|}$

$(9) \quad f(x)=e^x-2e^{-x}-3x$

$(10) \quad f(x)=\frac{\ln x}{\sqrt{x}}$

Exercise. Sketch a smooth curve $y=f(x)$ with $f(-2)=8,$ $f(0)=4,$ $f(2)=0,$ $f'(x)>0$ for $|x|>2,$ $f'(2)=f'(-2)=0,$ $f'(x)<0$ for $|x|<2,$ $f”(x)<0$ for $x<0,$ and $f”(x)>0$ for $x>0.$

Exercise. Sketch the graph of the following function $$f(x)=x^{2/3}(x-7).$$ Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

Exercise. Sketch the graph of the following function $$h(x)=\frac{3x-2}{\sqrt{2x^2+1}}.$$ Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

Exercise. Find constants $a$ and $b$ that guarantee that the graph of the function defined by $$f(x)=\frac{a x+5}{3-b x}$$ will have a vertical asymptote at $x=5$ and a horizontal asymptote at $y=-3.$ Sketch the graph of the function.

Exercise. Find all vertical tangents and cusps for the function $$f(x) = \sqrt{4-x^2}.$$ Justify your work. Sketch the graph of the function.

Exercise. Find all vertical tangents and cusps for the function \begin{equation} f(x) = \begin{cases} x^{1/3}+3 & x\leq 0 \\ 3-x^{1/5} & x\geq 0 \end{cases} \end{equation} Justify your work. Sketch the graph of the function.

Exercise. Explain why the function $$f(x)= \begin{cases} 1 & x\leq 0 \\ 1/x & x>0 \end{cases}$$ has a vertical asymptote but no vertical tangent. Sketch the graph of the function.

Exercise. Sketch the graph of the curve $$y^2=\frac{x^3}{2a-x}$$ for [$0,2a$). Show all special features such as vertical asymptotes, horizontal asymptotes, cusps, vertical tangents, and intercepts. Sketch the graph of the function.

Exercise. Sketch the graph of the following functions. Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

$(1) \quad f(x)=x^{2/3}(x-7)$

$(2) \quad g(x)=\dfrac{72-54 x+x^2+2 x^3}{15-17 x+4 x^2}$

$(3) \quad h(x)=\dfrac{3x-2}{\sqrt{2x^2+1}}$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.