Inflection Points

A curve's inflection point is the point at which the curve's concavity changes. For a function $f(x),$ its concavity can be measured by its second order derivative $f''(x).$ When $f''<0,$ which means that the function's rate of change is decreasing, the function is concave down. In contrast, when the function's rate of change is increasing, i.e. $f''>0,$ the function is concave up.

In typical problems, we find a function's inflection point by using $f''=0$ $($provided that $f$ and $f'$ are both differentiable at that point$)$ and checking the sign of $f''$ around that point. Be careful not to forget that $f''=0$ does not necessarily mean that the point is an inflection point since the sign of $f''$ might not change before and after that point.

In the figure above, the red zone depicts the area where the function is concave down and the blue zone indicates concave up.

What are the inflection points of the curve $y=x^4-4x^3-18x^2+15?$

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Let $y=f(x),$ then

$$\begin{aligned} f'(x)&=4x^3-12x^2-36x\\ f''(x)&=12x^2-24x-36\\ &=12(x+1)(x-3). \end{aligned}$$

Checking the signs of $f''(x)$ around $x=-1$ and $x=3,$ we get the table below:

$$\begin{array} { c c r c r c } x & \cdots & -1 & \cdots & 3 & \cdots \\ f''(x) & (+) & 0 & (-) & 0 & (+) \end{array}$$

This table tells us that $f(x)$ is concave up for $x<-1,$ concave down for $-1<x<3,$ and concave up for $x>3.$ Hence, the two inflection points of the curve $y=f(x)$ are $\big(-1, f(-1)\big)$ and $\big(3, f(3)\big),$ or equivalently,

$$(-1, 2),\ \ (3, -174). \ _\square$$

What is the slope of the tangent of the curve $y=x^3-6x^2+12x-7$ at its inflection point?

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Let $y=f(x),$ then

$$\begin{aligned} f'(x)&=3x^2-12x+12=3(x-2)^2\\ f''(x)&=6x-12=6(x-2). \end{aligned}$$

The values of $f'(x)$ and $f''(x)$ are both $0$ at $x=2.$ Checking the signs of $f'(x)$ and $f''(x)$ around $x=2,$ we get the table below:

$$\begin{array} { c c r c } x & \cdots & 2 & \cdots \\ f'(x) & (+) & 0 & (-) \\ f''(x) & (-) & 0 & (+) \end{array}$$

The swithcing signs of $f''(x)$ in the table tells us that $f(x)$ is concave down for $x<2$ and concave up for $x>2,$ implying that the point $\big(2, f(2)\big)=(2, 1)$ is the inflection point of the graph $y=f(x).$ Since the table also tells us that $f'(2)=0,$ the slope of the tangent of $f(x)$ at its inflection point $(2, 1)$ is $0.$ $_\square$

How many inflection points does $\sin x+\frac{1}{2}x^2$ have in the interval $[0,4\pi]?$

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Let $f(x)=\sin x+\frac{1}{2}x^2.$ Then, differentiating $f(x)$ twice gives

$$\begin{aligned} f(x)&=\sin x+\frac{1}{2}x^2\\ \Rightarrow f'(x)&=\cos x+x\\ \Rightarrow f''(x)&=-\sin x+1. \end{aligned}$$

Since $-1\leq\sin x\leq1,$ it is true that $0\leq-\sin x+1\leq2.$ Thus, $f''$ is either zero or positive, so the sign of $f''$ does not change. Therefore, $\sin x+\frac{1}{2}x^2$ has no inflection points in the interval $[0,4\pi].$ $_\square$

If $f'(x)=\frac{1}{4}x^4-\frac{7}{3}x^3+\frac{15}{2}x^2-9x+2,$ how many inflection points does the function $f(x)$ have?

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The second order derivative of $f(x)$ is

$$\begin{aligned} f'(x)&=\frac{1}{4}x^4-\frac{7}{3}x^3+\frac{15}{2}x^2-9x+2\\ \Rightarrow f''(x)&=x^3-7x^2+15x-9\\ &=(x-1)(x-3)^2. \end{aligned}$$

Thus, $f''=0$ at $x=1$ and $x=3.$ Checking the signs of $f''(x)$ around $x=1$ and $x=3,$ we get the table below:

$$\begin{array} { c c r c r c } x & \cdots & 1 & \cdots & 3 & \cdots \\ f''(x) & (-) & 0 & (+) & 0 & (+) \end{array}$$

Since the sign of $f''$ does not change before and after $x=3,$ the function only has an inflection point at $x=1.$ Therefore the answer is 1. $_\square$