- Home
- >
- Concavity calculus – Concave Up, Concave Down, and Points of Inflection

# Concavity calculus – Concave Up, Concave Down, and Points of Inflection

**Concavity in Calculus**helps us predict the shape and behavior of a graph at critical intervals and points. Knowing about the graphâ€™s concavity will also be helpful when sketching functions with complex graphs.

**Our discussion will focus on the following concepts and techniques:**

*Concavity calculus highlights the importance of the function’s second derivative in confirming whether its resulting curve concaves upward, downward, or is an inflection point at its critical points.*- Identifying concavity and points of inflections given a functionâ€™s graph.
- Applying the second derivative test to determine the concavity of a function at different critical points.
- Understanding the relationship between $f(x)$, $f^{\prime}(x)$, and $f^{\prime\prime} (x)$ and how it affects the functionâ€™s shape.

**What is concavity?**

Concavity tells us the shape and **how a function bends throughout its interval**. When given a functionâ€™s graph, observe the

**points where they concave downward or downward**. These will tell you the concavity present at the function.

**points where the curveâ€™s concavity changes**. We call these points

**inflection points**. The three examples shown above can guide you in identifying the concavity of a function given its curve as well.

- From this, we can see that when a curve is concaving upward, expect a local minimum at the critical point.
- Similarly, at the point where the curve concaves downward, lookout for a local maximum at the critical point.
- Observe how the inflection point is the turning point between these two concavities.

**How to determine the concavity of a function?**

Letâ€™s begin this section by observing these three graphs showing how the curves of $f(x)$, $f^{\prime} (x)$, and $f^{\prime\prime}(x)$ are behaving at the critical numbers, $x=\{0, 1, 2\}$.
- The sign of $ f^{\prime}(x)$ changes from positive to negative within the interval that contains $x=0$.
- Similarly, $ f^{\prime}(x)$â€™s signs change from negative to positive within the interval that contains $x=2$.
- When $x = 0$, $f^{\prime\prime}(x)$ is negative while $f^{\prime\prime}(x)$ is positive when $x = 2$.
- Focusing on $x=1$, we can see that $ f^{\prime}(x)$ is equal to 0 and so is $f^{\prime\prime}(x)$.

**Test for concavity**

When the function, $f(x)$, is continuous and twice differentiable, we can use its second derivative to confirm concavity.
- When $f^{\prime\prime}(x) >0$, the graph is concaving upward.
- When $f^{\prime\prime}(x) <0$, the graph is concaving downward.
- When $f^{\prime\prime}(x) = 0$, the graph has an inflection point.

- When $f^{\prime}(x)$â€™s sign changes from positive to negative, the graphâ€™s curve is concaving downward.
- When $f^{\prime}$â€™s sign changes from negative to positive, the graphâ€™s curve is concaving upward.

**find the second derivative**of the function right away.

**Equate the second derivative to zero**. Use the solutions to

**divide the functionâ€™s domain into smaller intervals**then find a

**test value to determine the functionâ€™s concavity**at the intervals. Letâ€™s say we have $f(x) = x^4 â€“ 4x^2$. To predict its curveâ€™s concavity, we differentiate $f(x)$ twice in a row as shown below. \begin{aligned}f(x)&=x^4 – 8x^2\\f^{\prime}&= 4x^{3-1} -8(2x^{2-1}),\phantom{x}\color{Teal} \text{Power Rule & Constant Multiple Rule}\\&=4x^3 -16x\\\\ f^{\prime\prime}(x) &=4(3x^{3-1})-16(x^{1 -1}) ,\phantom{x}\color{Teal} \text{Power Rule & Constant Multiple Rule}\\&=12x^2 – 16\end{aligned} Equate $f^{\prime\prime}(x)$ to 0 so that we can divide the domain of $f(x)$ into smaller intervals. \begin{aligned}f^{\prime\prime}(x)&=12x^2 – 16\\12x^2-16 &= 0\\12x^2&= 16\\x^2&= \dfrac{4}{3}\\x&=\pm\dfrac{2}{\sqrt{3}}\\&\approx 1.15\end{aligned} Letâ€™s use the approximate values of $x$ since we just need to find test values that are within the intervals: $\left(-\infty, -\dfrac{2}{\sqrt{3}}\right)$, $\left(-\dfrac{2}{\sqrt{3}},\dfrac{2}{\sqrt{3}}\right)$, and $\left(\dfrac{2}{\sqrt{3}}\right)$.

Interval |
Test Value |
Sign of $\boldsymbol{ f^{\prime\prime}(x)}$ |
Concavity |

\begin{aligned}(-\infty, -1.15)\end{aligned} | \begin{aligned} x&= -3\end{aligned} | \begin{aligned} f^{\prime\prime}(-3)&= 12(-3)^2 -16\\&= 92\\&\Rightarrow \color{Teal}\text{Positive}\end{aligned} | Concave upward |

\begin{aligned}\left(-1.15, 1.15\right)\end{aligned} | \begin{aligned} x&= 1\end{aligned} | \begin{aligned} f^{\prime\prime}(-1)&= 12(-1)^2 -16\\&= -4\\&\Rightarrow \color{Orchid}\text{Negative}\end{aligned} | Concave downward |

\begin{aligned} (1.15,\infty)\end{aligned} | \begin{aligned} x&= 3\end{aligned} | \begin{aligned} f^{\prime\prime}(3)&= 12(3)^2 -16\\&= 92\\&\Rightarrow \color{Teal}\text{Positive}\end{aligned} | Concave upward |

**How to compare concave up vs concave down?**

Letâ€™s summarize what weâ€™ve just learned throughout the discussion by comparing the curves that concave upward or downward.
**Concave up**

The upper half breaks down the behavior of $ f^{\prime}(x)$ and $ f^{\prime\prime}(x)$ when the curve is concaving upwards.
**second derivative is positive when the curve is concaving upward**. When the curve is concaving upward, the function is decreasing then increasing. This means that $\boldsymbol{f^{\prime}(x)}$

**changes from positive to negative**.

**Concave down**

The upper half tells us that when the curve is concaving downward, the **second derivative of the function at that point is negative**.

**the curve concaving downward is increasing from the left and decreasing from the right**.

**Inflection point**

We canâ€™t discuss concavity without summarizing what we know about inflection points.
**Inflection points**are located at values of $\boldsymbol{x}$

**the curve changes concavity.**Keep in mind that at this point, $\boldsymbol{f^{\prime\prime}(x)}$

**is equal to zero**. This section summarizes what weâ€™ve just learned about concavities. Review the example weâ€™ve shown you in our discussion too. Once youâ€™re ready, dive right into these sample problems weâ€™ve prepared just for you!

**Given $f(x) = x^3âˆ’24x^2+24$, find the open intervals where $f(x)$â€™s curve is concaving upward or downward.**

*Example 1*__Solution__We begin by differentiating $f(x)$ twice in row to determine the expression for $f^{\prime \prime}$. \begin{aligned}f(x)&=x^3- 12x^2 +24\\f^{\prime}(x)&=\dfrac{d}{dx}x^3 -\dfrac{d}{dx}12x^2+ \dfrac{d}{dx}24, \phantom{x}\color{Teal}\text{Sum & Difference Rules}\\&=\dfrac{d}{dx}x^3 -\dfrac{d}{dx}12x^2+ 0, \phantom{x}\color{Teal}\text{Constant Rule}\\&= 3x^{3-1} – 12(2x^{2-1}),\phantom{x}{\color{Teal} \text{Constant Multiple & Power Rules}}\\&= 3x^2-24x\\\\f^{\prime\prime}(x) &= 3(2x^{2 -1})-24(x^{1- 1}),\phantom{x}{\color{Teal} \text{Constant Multiple & Power Rules}}\\&=6x – 24\end{aligned} Now that we have $f^{\prime \prime}(x)$, equate the expression to $0$ then find $x$. Use the values of $x$ to divide the domain of $f(x)$ into smaller intervals. \begin{aligned}f^{\prime\prime}(x) &=0\\6x – 24&= 0\\6x &= 24\\x &=4\end{aligned} Observe the signs of $f^{\prime\prime}(x)$ for the intervals, $(-\infty, 4)$ and $(4, \infty)$. When $f^{\prime \prime}(x)$ is negative, $f(x)$ is concaving downward. Similarly, when $f^{\prime \prime}(x)$ is positive, $f(x)$ is concaving upward.

Interval |
Test Value |
Sign of $\boldsymbol{ f^{\prime\prime}(x)}$ |
Concavity |

\begin{aligned}(-\infty, 4)\end{aligned} | \begin{aligned} x&= -4\end{aligned} | \begin{aligned} f^{\prime\prime}(1)&= 6(1) -24 \\&=-18\\&\Rightarrow \color{Orchid}\text{Negative}\end{aligned} | Concave downward |

\begin{aligned}(4,\infty)\end{aligned} | \begin{aligned} x&= 4\end{aligned} | \begin{aligned} f^{\prime\prime}(5)&= 6(5) â€“ 24\\&=6\\&\Rightarrow \color{Teal}\text{Positive}\end{aligned} | Concave upward |

**Given $f(x) = 3x^2 – \dfrac{1}{x^2}$, find its points of inflection. Discuss the concavity of the functionâ€™s graph as well. Â Form the acquired information, roughly sketch the graph of $f(x)$.**

*Example 2*__Solution__Recall that $\dfrac{1}{x^2} = x^{-2}$, so we can rewrite $f(x)$ first to differentiate it much faster. Find $f^{\prime \prime}(x)$ by differentiating $f(x)$ twice in a row as shown below. \begin{aligned}f(x) &= 3x^2 – \dfrac{1}{x^2}\\&= 3x^2 – x^{-2}\\f^{\prime}(x) &= 3(2x^{2 -1}) – (-2)x^{-2 -1},\phantom{x}\color{Teal}\text{Constant Multiple & Power Rules}\\&= 6x + 2x^{-3}\\f^{\prime \prime}(x)&= 6(x^{1-1})+2(-3x^{-3 -1}),\phantom{x}\color{Teal}\text{Constant Multiple & Power Rules}\\&=6x^0 – 6x^{-4}\\&= 6 – \dfrac{6}{x^4}\end{aligned} Equate $f^{\prime\prime}(x)$ to zero then divide the domain of $f(x) = 3x^2 – \dfrac{1}{x^2}$ into smaller intervals. \begin{aligned}f^{\prime \prime}(x)&=0\\6 – \dfrac{6}{x^4}&= 0\\\dfrac{1}{x^4} &= 1\\x&= \pm 1\end{aligned} The domain of $f(x)$ is $(-\infty, 0) \cup (0, \infty)$, so we can divide this into smaller intervals to account for $x = \pm 1$. Assign test values that are within the following intervals: $\{(-\infty, -1), (-1, 0), (0, 1), (1, \infty)\}$.

Interval |
Test Value |
Sign of $\boldsymbol{ f^{\prime\prime}(x)}$ |
Concavity |

\begin{aligned}(-\infty, -1)\end{aligned} | \begin{aligned} x&= -2\end{aligned} | \begin{aligned} f^{\prime\prime}(-2)&= 6 – \dfrac{6}{(-2)^4}\\&=\dfrac{45}{8}\\&\Rightarrow \color{Teal}\text{Positive}\end{aligned} | Concave upward |

\begin{aligned}(-1,0)\end{aligned} | \begin{aligned} x&= -\dfrac{1}{2}\end{aligned} | \begin{aligned} f^{\prime\prime}\left(-\dfrac{1}{2} \right )&= 6 – \dfrac{6}{\left(-\dfrac{1}{2} \right )^4}\\&=-90\\&\Rightarrow \color{Orchid}\text{Negative}\end{aligned} | Concave downward |

\begin{aligned}(0, 1)\end{aligned} | \begin{aligned} x&= \dfrac{1}{2}\end{aligned} | \begin{aligned} f^{\prime\prime}\left(\dfrac{1}{2} \right )&= 6 – \dfrac{6}{\left(\dfrac{1}{2} \right )^4}\\&=-90\\&\Rightarrow \color{Orchid}\text{Negative}\end{aligned} | Concave downward |

\begin{aligned}(1,\infty)\end{aligned} | \begin{aligned} x&= 2\end{aligned} | \begin{aligned} f^{\prime\prime}(2)&= 6 – \dfrac{6}{(2)^4}\\&=\dfrac{45}{8}\\&\Rightarrow \color{Teal}\text{Positive}\end{aligned} | Concave upward |

**Practice Questions**

1. Given the functions shown below, find the open intervals where each functionâ€™s curve is concaving upward or downward.
a. $f(x) = x^3 â€“ 12x + 18$
b. $g(x) = \dfrac{1}{4}x^4 â€“ \dfrac{1}{3}x^3 + \dfrac{1}{2}x^2$
c. $h(x) =x^5 â€“ 270x^2 + 1$
2. Given the functions shown below, find the open intervals where each functionâ€™s curve is concaving upward or downward.
a. $f(x) = \dfrac{x}{x+1}$
b. $g(x) = x\sqrt{x^2 -1}$
c. $h(x) =4x^2 – \dfrac{1}{x}$
3. Given $f(x) = 2x^4 â€“ 4x^3$, find its points of inflection. Discuss the concavity of the functionâ€™s graph as well. Form the acquired information, roughly sketch the graph of $f(x)$.
**Answer Key**

1.
a. Concave downward: $(-\infty, 0)$; Concave upward: $(0, \infty)$
b. Concave upward throughout the domain – $(-\infty, \infty)$.
c. Concave downward: $(-\infty, 3)$; Concave upward: $(3, \infty)$
2.
a. Concave downward: $(-\infty, -1)$; Concave upward: $(-1, \infty)$
b. Concave downward: $\left(-\infty, -\sqrt{\dfrac{3}{2}}\right)$ and $\left(1,\sqrt{\dfrac{3}{2}}\right)$; Concave upward: $\left(-\sqrt{\dfrac{3}{2}}, -1\right)$ and $\left(\sqrt{\dfrac{3}{2}}, \infty\right)$
c. Concave downward: $\left(0, \dfrac{\sqrt[3]{2}}{2}\right)$; Concave upward: $\left(-\infty, 0\right)$ and $\left(\dfrac{\sqrt[3]{2}}{2}, \infty\right)$
3.
Concavity: Concave downward: $(0, 1)$; Concave upward: $(-\infty, 0)$ and $(1, \infty)$
Points of Inflection: $(0,0)$ and $(1, -2)$
Graph:
*Images/mathematical drawings are created with GeoGebra.*

Rate this page