# Calculus: Derivatives III

## Inflection Points

An inflection point is a point where the concavity of a function transitions from concave up to concave down, or from concave down to concave up. These points occur when the second derivative is either zero or undefined (think of a piecewise function). In order to find all the inflection points of a function, simply find all the points in the second derivative that are either zero or undefined. From there, make sure that the second derivative does in fact transition from positive to negative (or vice versa) by testing a point in the region between such points (or between the point and $\pm\infty$ at the edges).

## Problems

1. Find the inflection points of the following function: $y = x^3$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx} x^3 \\ \dfrac{dy}{dx} = 3x^2 \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}3x^2 \\ \dfrac{d^2y}{dx^2} = 6x \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ 6x = 0 \\ x = 0 \\$

Step 3: Determine whether the zero is an inflection point:

$\dfrac{d^2y}{dx^2}(-1)=6(-1) \\ \dfrac{d^2y}{dx^2}(-1)=-6 \\$

$\dfrac{d^2y}{dx^2}(1)=6(1) \\ \dfrac{d^2y}{dx^2}(1)=6 \\$

The second derivative changes from negative to positive when $x=0$, therefore the function has an inflection point when $x=0$.

Step 4: Calculate the inflection point's $y$-value:

$y = 0^3 \\ y = 0 \\$

The inflection point is at $(0,0)$.

2. Find the inflection points of the following function: $y = 2x^3 -x^2 + 8x + 3$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx} ( 2x^3 -x^2 + 8x + 3 ) \\ \dfrac{dy}{dx} = 6x^2 - 2x + 8\\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} (6x^2 - 2x + 8) \\ \dfrac{d^2y}{dx^2} = 12x - 2 \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ 12x - 2 = 0 \\ 12x = 2 \\ x = \dfrac{1}{6} \\$

Step 3: Determine whether the zero is an inflection point:

$\dfrac{d^2y}{dx^2}(0) = 12(0)-2 \\ \dfrac{d^2y}{dx^2}(0) = -2 \\$

$\dfrac{d^2y}{dx^2}(1) = 12(1) - 2 \\ \dfrac{d^2y}{dx^2}(1) = 10 \\$

The second derivative changes from negative to positive when $x = \dfrac{1}{6}$, therefore the function has an inflection point when $x=\dfrac{1}{6}$.

Step 4: Calculate the inflection point's $y$-value:

$y = 2\left(\dfrac{1}{6}\right)^3 - \left(\dfrac{1}{6}\right)^2 + 8\left(\dfrac{1}{6}\right) + 3 \\ y = \dfrac{2}{216} - \dfrac{1}{36} + \dfrac{8}{6} + 3 \\ y = \dfrac{233}{54} \\$

The inflection point is at $\left(\dfrac{1}{6},\dfrac{233}{54}\right)$.

3. Find the inflection points of the following function: $y = x^4-6x^2 + 3x - 10$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx} ( x^4-6x^2 + 3x - 10 ) \\ \dfrac{dy}{dx} = 4x^3-12x + 3 \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} (4x^3-12x + 3) \\ \dfrac{d^2y}{dx^2} = 12x^2-12 \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ 12x^2 - 12 = 0 \\ x^2-1 = 0 \\ (x+1)(x-1) = 0 \\ x = 1, \, x = -1$

Step 3: Determine whether the zeros are inflection points:

$\dfrac{d^2y}{dx^2}(-2) = 12(-2)^2 - 12 \\ \dfrac{d^2y}{dx^2}(-2) = 12(4) - 12 \\ \dfrac{d^2y}{dx^2}(-2) = 36 \\$

$\dfrac{d^2y}{dx^2}(0) = 12(0)^2 - 12 \\ \dfrac{d^2y}{dx^2}(0) = - 12 \\$

$\dfrac{d^2y}{dx^2}(2) = 12(2)^2 - 12 \\ \dfrac{d^2y}{dx^2}(2) = 12(4) - 12 \\ \dfrac{d^2y}{dx^2}(2) = 36 \\$

The second derivative changes from positive to negative when $x=-1$, and from negative to positive when $x=1$, therefore the function has inflection points at both.

Step 4: Calculate the inflection point's $y$-value:

$y = 1^4 - 6\cdot 1^2 + 3\cdot 1 - 10 \\ y = 1 - 6 + 3 - 10 \\ y = -12 \\ \, \\ y = (-1)^4 - 6(-1)^2 + 3(-1) - 10 \\ y = 1 - 6 - 3 - 10 \\ y = -18$

The inflection points are at $(1,-12)$ and $(-1,-18)$.

4. Find the inflection points of the following function: $y = e^{x} + x^2$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx} ( e^x + x^2 ) \\ \dfrac{dy}{dx} = e^x + 2x \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} ( e^x + 2x) \\ \dfrac{d^2y}{dx^2} = e^x + 2 \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ e^x + 2 = 0 \\$

The second derivative is defined everywhere and has no zeros, therefore the function has no inflection points.

5. Find the inflection points of the following function: $y=x^4 + 6x^3 + 12x^2 - 4x + 1$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}(x^4 + 6x^3 + 12x^2 - 4x + 1\\ \dfrac{dy}{dx} = 4x^3 + 18x^2 + 24x - 4\\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}(4x^3 + 18x^2 + 24x - 4) \\ \dfrac{d^2y}{dx^2} = 12x^2 + 36x + 24 \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ 12x^2 + 36x + 24 = 0 \\ x^2 + 3x + 2 = 0 \\ (x+1)(x+2) = 0 \\ x=-2, x=-1$

Step 3: Determine whether the zeros are inflection points:

$\dfrac{d^2y}{dx^2}(-2) = 12(-3)^2 + 36(-3) + 24 \\ \dfrac{d^2y}{dx^2}(-2) = 12(9) + 36(-3) + 24 \\ \dfrac{d^2y}{dx^2}(-2) = 108 - 108 + 24 \\ \dfrac{d^2y}{dx^2}(-2) = 24 \\$

$\dfrac{d^2y}{dx^2}(-1.5) = 12(-1.5)^2 + 36(-1.5) + 24 \\ \dfrac{d^2y}{dx^2}(-1.5) = 12(2.25) + 36(-1.5) + 24 \\ \dfrac{d^2y}{dx^2}(-1.5) = 27 - 54 + 24 \\ \dfrac{d^2y}{dx^2}(-1.5) = -3 \\$

$\dfrac{d^2y}{dx^2}(0) = 12(0)^2 + 36(0) + 24 \\ \dfrac{d^2y}{dx^2}(0) = 24 \\$

The second derivative changes from positive to negative when it crosses $x=-2$ and from negative to positive when it crosses $x=-1$. Therefore both zeros are inflection points.

Step 4: Calculate the $y$-values of the inflection points:

$y = (-2)^4 + 6(-2)^3 + 12(-2)^2 - 4(-2) + 1 \\ y = 16 + 6(-8) + 12(4) + 8 + 1 \\ y = 16 - 48 + 48 + 8 + 1 \\ y = 25 \\$

$y = (-1)^4 + 6(-1)^3 + 12(-1)^2 - 4(-1) + 1 \\ y = 1 - 6 + 12 + 4 + 1 \\ y = 12 \\$

The inflection points are at $(-2,25)$ and $(-1,12)$.

6. Find the inflection points of the following function: $y = \sin(\theta)$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx} \sin(\theta) \\ \dfrac{dy}{dx} = \cos(\theta) \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}\cos(\theta) \\ \dfrac{d^2y}{dx^2} = -\sin(\theta) \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ -\sin(\theta) = 0 \\ \sin(\theta) = 0 \\ \theta = \sin^{-1}(0) \\ \theta = k\pi, \, \, \text{where } \, k \, \text{ is an integer}$

Step 3: Determine whether the zeros are inflection points:

Since sine is periodic, we only need to test points in $[0,2\pi]$ to determine concavity.

$\dfrac{d^2y}{dx^2}\left(\dfrac{\pi}{2}\right) = -\sin\left(\dfrac{\pi}{2}\right) \\ \dfrac{d^2y}{dx^2}\left(\dfrac{\pi}{2}\right) = -1 \\$

$\dfrac{d^2y}{dx^2}\left(\dfrac{3\pi}{2}\right) = -\sin\left(\dfrac{3\pi}{2}\right) \\ \dfrac{d^2y}{dx^2}\left(\dfrac{3\pi}{2}\right) = -(-1) \\ \dfrac{d^2y}{dx^2}\left(\dfrac{3\pi}{2}\right) = 1 \\$

The second derivative changes from negative to positive across $x=\pi$, and then back from positive to negative across $x=0=2\pi$. Therefore, the zeros of the second derivative are all inflection points.

Step 4: Calculate the inflection point's $y$-value:

$y = \sin(k\pi) \\ y = 0 \\$

The inflection points are at $(k\pi,0)$, where $k$ is an integer.

7. Find the inflection points of the following function: $y=\tan(\theta)$.

Step 1: Take the second derivative:

$\dfrac{d}{d\theta}y = \dfrac{d}{d\theta}\tan(\theta) \\ \dfrac{dy}{d\theta} = \sec^2(\theta) \\ \dfrac{d}{d\theta}\dfrac{dy}{d\theta} = \dfrac{d}{d\theta} \sec^2(\theta) \\ \dfrac{d^2y}{d\theta^2} = \sec^2(\theta)\tan(\theta) \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{d\theta^2} = 0 \\ \sec^2(\theta)\tan(\theta) = 0 \\$

The second derivative is zero when either $\sec^2(\theta)=0$ or when $\tan(\theta)=0$. We see that $\sec^2(\theta)$ is always positive, and $\tan(\theta)=0$ when $\theta = k\pi$, where $k \in \mathbb{Z}$. Therefore the zeros of the second derivative are at when $\theta=k\pi$, where $k \in \mathbb{Z}$.

Step 3: Determine whether the zeros are inflection points:

$\dfrac{dy^2}{d\theta^2}\left(\dfrac{-\pi}{4}\right) = \sec^2\left(\dfrac{-\pi}{4}\right)\tan\left(\dfrac{-\pi}{4}\right) \\ \dfrac{dy^2}{d\theta^2}\left(\dfrac{-\pi}{4}\right) = (2)(-1) \\ \dfrac{dy^2}{d\theta^2}\left(\dfrac{-\pi}{4}\right) = -2 \\$

$\dfrac{dy^2}{d\theta^2}\left(\dfrac{\pi}{4}\right) = \sec^2\left(\dfrac{\pi}{4}\right)\tan\left(\dfrac{\pi}{4}\right) \\ \dfrac{dy^2}{d\theta^2}\left(\dfrac{\pi}{4}\right) = (2)(1) \\ \dfrac{dy^2}{d\theta^2}\left(\dfrac{\pi}{4}\right) = 2 \\$

The function transitions from concave down to concave up when $\theta=k\pi$, where $k \in \mathbb{Z}$. Because the function is undefined when $\theta = \dfrac{\pi}{2}+k\pi$, where $k \in \mathbb{Z}$, the zeros are all inflection points.

Step 4: Calculate the inflection points' $y$-values:

Because $\tan(\theta)$ is periodic, we only need to test the value $\theta=0$.

$\tan(0) = 0$

The inflection points are located at $(k\pi,0)$, where $k \in \mathbb{Z}$.

8. Find the inflection points of the following function: $y= 12\ln(x)$ where $x > 0$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}12\ln(x) \\ \dfrac{dy}{dx} = \dfrac{12}{x} \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} \dfrac{12}{x} \\ \dfrac{d^2y}{dx^2} = \dfrac{-12}{x^2} \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ \dfrac{-12}{x^2} = 0 \\$

The second derivative is defined everywhere and has no zeros, therefore the function has no inflection points.

9. Find the inflection points of the following function: $y = \dfrac{x}{x^2+1}$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{x}{x^2+1} \\ \dfrac{dy}{dx} = \dfrac{(x^2+1)(1) - x(2x)}{(x^2+1)^2}\\ \dfrac{dy}{dx} = \dfrac{-x^2 + 1}{(x^2+1)^2}\\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx} \dfrac{-x^2 + 1}{(x^2+1)^2} \\ \dfrac{d^2y}{dx^2} = \dfrac{(x^2+1)^2(-2x) - (-x^2+1)(2(x^2+1)(2x))}{(x^2+1)^4}\\ \dfrac{d^2y}{dx^2} = \dfrac{2x(x^2+1)(x^2-3)}{(x^2+1)^4}\\ \dfrac{d^2y}{dx^2} = \dfrac{2x(x^2-3)}{(x^2+1)^3}\\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ \dfrac{2x(x^2-3)}{(x^2+1)^3} = 0 \\$

The denominator is never $0$, so we can safely restrict our attention to finding zeros in the numerator:

$2x(x^2-3) = 0 \\ 2x(x+\sqrt{3})(x-\sqrt{3}) \\ x=-\sqrt{3}, x=0, x=\sqrt{3} \\$

Step 3: Determine whether the zeros are inflection points:

Since the denominator of the second derivative is always positive, we can disregard it for ease of computation.

$\dfrac{dy^2}{dx^2}(-2) = 2(-2)((-2)^2-3) \\ \dfrac{dy^2}{dx^2}(-2) = (-4)(4-3) \\ \dfrac{dy^2}{dx^2}(-2) = (-4)(1) \\ \dfrac{dy^2}{dx^2}(-2) = -4 \\$

$\dfrac{dy^2}{dx^2}(-1) = 2(-1)((-1)^2-3) \\ \dfrac{dy^2}{dx^2}(-1) = (-2)(1-3) \\ \dfrac{dy^2}{dx^2}(-1) = (-2)(-2) \\ \dfrac{dy^2}{dx^2}(-1) = 4 \\$

$\dfrac{dy^2}{dx^2}(1) = 2(1)((-1)^2-3) \\ \dfrac{dy^2}{dx^2}(1) = (2)(1-3) \\ \dfrac{dy^2}{dx^2}(1) = (2)(-2) \\ \dfrac{dy^2}{dx^2}(1) = -4 \\$

$\dfrac{dy^2}{dx^2}(2) = 2(2)(2^2-3) \\ \dfrac{dy^2}{dx^2}(2) = (4)(4-3) \\ \dfrac{dy^2}{dx^2}(2) = (4)(1) \\ \dfrac{dy^2}{dx^2}(2) = 4 \\$

The second derivative changes signs each time it crosses a zero, therefore all the zeros are inflection points.

Step 4: Calculate the $y$-values of the inflection points:

$y = \dfrac{-\sqrt{3}}{(-\sqrt{3})^2+1} \\ y = \dfrac{-\sqrt{3}}{3+1} \\ y = \dfrac{-\sqrt{3}}{4} \\$

$y = \dfrac{0}{0^2+1} \\ y = 0$

$y = \dfrac{\sqrt{3}}{\sqrt{3}^2+1} \\ y = \dfrac{\sqrt{3}}{3+1} \\ y = \dfrac{\sqrt{3}}{4} \\$

The inflection points of the function are at $\left(-\sqrt{3}, \dfrac{-\sqrt{3}}{4}\right)$, $(0,0)$, and $\left(\sqrt{3}, \dfrac{\sqrt{3}}{4}\right)$.

10. Find the inflection points of the following function: $y=\dfrac{1}{x^2+1}$.

Step 1: Take the second derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{1}{x^2+1} \\ \dfrac{dy}{dx} = \dfrac{-2x}{(x^2+1)^2} \\ \dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{-2x}{(x^2+1)^2} \\ \dfrac{d^2y}{dx^2} = \dfrac{-2(x^2+1)^2 + 2x(2(x^2+1)(2x))}{(x^2+1)^4} \\ \dfrac{d^2y}{dx^2} = \dfrac{ 2(x^2+1)(3x^2-1)}{(x^2+1)^4} \\ \dfrac{d^2y}{dx^2} = \dfrac{ 2(3x^2-1)}{(x^2+1)^3} \\$

Step 2: Find the zeros of the second derivative:

$\dfrac{d^2y}{dx^2} = 0 \\ \dfrac{ 2(3x^2-1)}{(x^2+1)^3} = 0 \\$

Since the denominator is always positive, we can safely restrict our attention to the numerator. The zeros of the quadratic expression are at $x=-\dfrac{1}{\sqrt{3}}$ and $x=\dfrac{1}{\sqrt{3}}$.

Step 3: Determine whether the zeros are inflection points:

Since the denominator of the the second derivative is always positive, we can leave it out when calculating the sign.

$\dfrac{d^2y}{dx^2}(-2) = 2(3(2)^2-1) \\ \dfrac{d^2y}{dx^2}(-2) = 2(3(4)-1) \\ \dfrac{d^2y}{dx^2}(-2) = 22 \\$

$\dfrac{d^2y}{dx^2}(0) = 2(3(0)^2-1) \\ \dfrac{d^2y}{dx^2}(0) = 2(3(0)-1) \\ \dfrac{d^2y}{dx^2}(0) = -2 \\$

$\dfrac{d^2y}{dx^2}(-2) = 2(3(2)^2-1) \\ \dfrac{d^2y}{dx^2}(-2) = 2(3(4)-1) \\ \dfrac{d^2y}{dx^2}(-2) = 22 \\$

The second derivative changes from positive to negative as it crosses $x=-\dfrac{1}{\sqrt{3}}$ and then back to positive as it crosses $x=\dfrac{1}{\sqrt{3}}$. Therefore, both zeros are inflection points.

Step 4: Calculate the inflection points' $y$-values:

$y=\dfrac{1}{\left(-\dfrac{1}{\sqrt{3}}\right)^2+1} \\ y=\dfrac{1}{\dfrac{1}{3}+1} \\ y=\dfrac{1}{\dfrac{4}{3}} \\ y=\dfrac{3}{4} \\$

$y=\dfrac{1}{\left(\dfrac{1}{\sqrt{3}}\right)^2+1} \\ y=\dfrac{1}{\dfrac{1}{3}+1} \\ y=\dfrac{1}{\dfrac{4}{3}} \\ y=\dfrac{3}{4} \\$

The inflection points are at $\left(\dfrac{1}{\sqrt{3}}, \dfrac{3}{4}\right)$ and $\left(-\dfrac{1 }{\sqrt{3}}, \dfrac{3}{4}\right)$.