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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 graph’s concavity will also be helpful when sketching functions with complex graphs.
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.
Our discussion will focus on the following concepts and techniques:
- Identifying concavity and points of inflections given a function’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 function’s shape.
Review your derivative rules as well since we’ll have to apply them when finding the expressions of $ f^{\prime}(x)$ and $f^{\prime\prime}(x)$. We’ll assume knowledge of the first and second derivatives throughout the discussion, so make sure to take a quick refresher when you need to.
What is concavity?
Concavity tells us the shape and how a function bends throughout its interval. When given a function’s graph, observe the points where they concave downward or downward. These will tell you the concavity present at the function.
It’s also possible to find points where the curve’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.
Here’s an example of a graph that exhibits all three concavity types.
- 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.
In the next section, we’ll learn how to apply the second derivative test to confirm the concavity of a function given its expression.
How to determine the concavity of a function?
Let’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, $x=\{0, 1, 2\}$.
From $f(x)$’s graph, we can see that $x = 0$ is a relative maximum and the curve is concaving upward. The point at $x = 1$ is an inflection point while $x =2$ is a relative minimum. The graph also concaves downward at $x = 2$.
Now, let’s observe $ f^{\prime}(x)$ and $f^{\prime\prime}(x)$’s graphs:
- The sign of $ f^{\prime}(x)$ changes from positive to negative within the interval that contains $x=0$.
- Similarly, $ f^{\prime}(x)$’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 $f^{\prime\prime}(x)$.
These observations show us that when the function is concaving downward at a critical point, $x=c$, the second derivative is negative. Similarly, when the function’s curve is concaving upward, $f^{\prime\prime}(x)$ is positive.
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.
What if we can’t differentiate$f(x)$ twice in a row? Use the observations we had about the function’s first derivative instead.
- When $f^{\prime}(x)$’s sign changes from positive to negative, the graph’s curve is concaving downward.
- When $f^{\prime}$’s sign changes from negative to positive, the graph’s curve is concaving upward.
When given a function, find the second derivative of the function right away. Equate the second derivative to zero. Use the solutions to divide the function’s domain into smaller intervals then find a test value to determine the function’s concavity at the intervals.
Let’s say we have $f(x) = x^4 – 4x^2$. To predict its curve’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}
Let’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 |
Our results show that the curve of $f(x)$ is concaving downward at the interval, $\left(-\dfrac{2}{\sqrt{3}},\dfrac{2}{\sqrt{3}}\right)$. Meanwhile, the function’s curve is concaving upward at the intervals, $\left(-\infty, -\dfrac{2}{\sqrt{3}}\right)$ and $\left(\dfrac{2}{\sqrt{3}}\right)$.
The graph above shows the curve of $f(x)$ and confirms its concavity. We’ve included the inflection points and their values can be confirmed by finding the function’s value at $x = \pm \dfrac{2}{\sqrt{3}}$.
\begin{aligned}f\left(\pm\dfrac{2}{\sqrt{3}} \right )&= \left(\pm\dfrac{2}{\sqrt{3}}\right)^4 -8\left(\pm\dfrac{2}{\sqrt{3}} \right )^2\\&=-\dfrac{80}{9}\\&\Rightarrow \left(-\dfrac{2}{\sqrt{3}}, -\dfrac{80}{9} \right ),\left(\dfrac{2}{\sqrt{3}}, -\dfrac{80}{9} \right )\end{aligned}
This means that $f(x)$ has inflection points at $\left(-\dfrac{2}{\sqrt{3}}, -\dfrac{80}{9} \right )$ and $\left(\dfrac{2}{\sqrt{3}}, -\dfrac{80}{9} \right )$.
How to compare concave up vs concave down?
Let’s summarize what we’ve just learned throughout the discussion by comparing the curves that concave upward or downward.
The four curves below will guide you in knowing the curve’s behavior given its first and second derivatives at the point. Let’s use these to highlight the difference between a curve that concaves 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.
This shows that the function’s 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 first derivative of the function is positive from the left and negative from the right of the critical point. This means that the curve concaving downward is increasing from the left and decreasing from the r