## What is the Concavity of a Function?

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the first derivative test and second derivative test to understand the concave behavior of the function.

## How to Calculate Function’s Concavity?

The calculation of a function's concavity is done using the second derivative test.

- The rate of change of a function's derivative is referred to as concavity. Where the derivative
*f*' is increasing, a function*f*is concave up (or upwards). The increase in the derivative of*f**'*is the same as saying that*f*" is greater than or equal to 0. Similarly, where the derivative*f*' is decreasing (or equivalently,*f*" is negative),*f*is concave down (or downwards). These are known as the first derivative test and second derivative test.

- In calculus, the path of curvature is not revealed by the sign of the derivative of f, which shows where the graph of
*f*is increasing or decreasing. For example, when the graph can be increasing on both sides of the point, but it has an upward curvature on the left side (“holds water”) and a downward curvature on the right side (“spills water”). - The maxima of smooth functions are found where the function is concave downward, while the minima are found where the function is concave upward.

### Geometrical Interpretation

- In calculus, we assume
*f*is concave up on intervals where the graph of*f*has upward curvature, and we say*f*is concave down on intervals where the graph has downward curvature. - If the tangents to a function f on an open interval have increasing slopes,
*f*is concave up; and if they have decreasing slopes, then f it is concave down. - On an open interval,
*f*is concave up if its graph is above its tangent lines, and concave down if it is below its tangent lines.

## Inflection Point

In differential calculus, an inflection point, a point of inflection, flex, or emphasis (intonation) is a point on a bend at which the bend changes from being inward (curved descending) to raised (sunken upward), or the other way around. A point where the curve disappears however doesn't change the sign is called a state of undulation or undulation point. In logarithmic calculation, an inflection point is characterized as a point where the digression meets the bend to arrange in any event, and an undulation point or hyperflex is characterized as a point where the digression meets the bend to arrange at any rate.

The inflection point is calculated with the help of a second derivative test. The point where the concave nature of the graph changes is known as an inflection point.

## Concavity using a Graphical Approach

- Consider the function $f(x)={x}^{3}+1$. It is a cubic function. Without using the first derivative test and second derivative test we can determine the concave nature of the function.
- The graph is of cup shape on the interval $\left(0,\infty \right)$ and is of cap shape on the interval $\left(-\infty ,0\right)$.
- The function is concave up on the interval $\left(0,\infty \right)$ and concave down on the interval $\left(-\infty ,0\right)$.
- The concave nature of the function changes at $\left(0,1\right)$. The point $\left(0,1\right)$ is an inflection point.

### Mathematical Justification

Mathematically, the statement can be justified using the first derivative test and second derivative test.

- Differentiate the function $f(x)={x}^{3}+1$ with respect to $x$ two times.
- $\begin{array}{l}f(x)={x}^{3}+1\\ f\text{'}\left(x\right)=3{x}^{2}\\ f\text{'}\text{'}\left(x\right)=6x\end{array}$
- So, for all $x>0$, the second derivative of the function $f\text{'}\text{'}\left(x\right)>0$ and for all $x<0$, the second derivative of the function $f\text{'}\text{'}\left(x\right)<0$.
- Therefore, the function is concave up on $\left(0,\infty \right)$ and concave down on the interval $\left(-\infty ,0\right)$. So, the function shows a concave behavior.
- The inflection point is given by $x=0$ as the function changes its concave nature at $x=0$.

## Practice Problems

### Problem 1

Consider the provided function $f\left(x\right)=10{x}^{2}-50x$. Using the first derivative test and second derivative test, determine the intervals in which function is concave down and concave up. Also, find the inflection point that is the point where the concavity changes.

**Solution:**

Evaluate the first derivative of the function,

Apply sum rule of differentiation,

$\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(10{x}^{2}-50x\right)\\ =\frac{d}{dx}\left(10{x}^{2}\right)+\frac{d}{dx}\left(-50x\right)\end{array}$

Apply the power rule of differentiation, $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$.

$\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(10{x}^{2}-50x\right)\\ =\frac{d}{dx}\left(10{x}^{2}\right)+\frac{d}{dx}\left(-50x\right)\\ =10\cdot 2{x}^{2-1}-50\\ =20x-50\end{array}$

To evaluate the second derivative of the function, differentiate the first derivative again with respect to *x*.

Apply the power rule of differentiation, $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$.

$\begin{array}{c}\frac{{d}^{2}}{d{x}^{2}}f\left(x\right)=\frac{d}{dx}\left(20x-50\right)\\ =\frac{d}{dx}\left(20x\right)-\frac{d}{dx}\left(50\right)\\ =20-0\\ =20\end{array}$

Observe that first derivative of the function $f\text{'}\left(x\right)=20x-50$.

Now, the first derivative can be both less and greater than 0, depending on values of *x*.

The first derivative will be negative if,

$\begin{array}{c}20x-50<0\\ 20x<50\\ x<\frac{50}{20}\\ x<2.5\end{array}$Therefore, the function $f\left(x\right)=10{x}^{2}-50x$is decreasing when $x<2.5$.

The first derivative will be greater than or equal to 0, if,

$\begin{array}{c}20x-50\ge 0\\ 20x\ge 50\\ x\ge \frac{50}{20}\\ x\ge 2.5\end{array}$Therefore, the function $f\left(x\right)=10{x}^{2}-50x$is increasing when $x\ge 2.5$.

Next, observe that the second derivative of the function $f\text{'}\text{'}\left(x\right)=20$is greater than 0 for all real values of x, so $f\left(x\right)=10{x}^{2}-50x$is concave up.

The function doesn’t change its concavity so no point of inflection.

### Problem 2

Consider the function $f\left(x\right)=x-{x}^{2}$. Evaluate the first derivative of the function. Find the interval where the function $f\left(x\right)=x-{x}^{2}$ is concave up and concave down. Check whether concavity changes or not.

**Solution:** Apply sum rule of differentiation,

$\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x-{x}^{2}\right)\\ =\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-{x}^{2}\right)\end{array}$

Apply the power rule of differentiation, $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$.

$\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x-{x}^{2}\right)\\ =\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-{x}^{2}\right)\\ =1-2x\end{array}$

To evaluate the second derivative of the function, differentiate the first derivative again with respect to *x*.

Apply the power rule of differentiation, $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$.

$\begin{array}{c}\frac{{d}^{2}}{d{x}^{2}}f\left(x\right)=\frac{d}{dx}\left(1-2x\right)\\ =\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(2x\right)\\ =0-2\\ =-2\end{array}$

Observe that first derivative of the function $f\text{'}\left(x\right)=1-2x$.

Now, the first derivative can be both negative and greater than or equal to 0 depending on values of *x*.

The first derivative will be negative if,

$\begin{array}{c}1-2x<0\\ 1<2x\\ \frac{1}{2}<x\\ x>0.5\end{array}$Therefore, the function $f\left(x\right)=x-{x}^{2}$ is decreasing when $x>0.5$.

The first derivative will be greater than or equal to 0 if,

$\begin{array}{c}1-2x\ge 0\\ 1\ge 2x\\ \frac{1}{2}\ge x\\ x\le 0.5\end{array}$Therefore, the function $f\left(x\right)=x-{x}^{2}$ is increasing when $x\le 0.5$.

Next, observe that the second derivative of the function $f\text{'}\text{'}\left(x\right)=-2$ is negative for all real values of x, so $f\left(x\right)=x-{x}^{2}$ is concave down on the entire real number line.

The function does not change its concavity so there is no inflection point.

### Problem 3

Find the interval where the function $f\left(x\right)={x}^{4}+2$ is concave up and concave down. Check whether concavity changes or not.

**Solution:**

- Consider the given function. $f\left(x\right)={x}^{4}+2$.
- We differentiate the function with respect to $x$ two times to obtain $f\text{'}\text{'}\left(x\right)=12{x}^{2}$.
- For all x, the second derivative $f\text{'}\text{'}\left(x\right)=12{x}^{2}$ is strictly greater than 0.

- The function is concave up on the entire real number line. For the function to be concave down, the double derivative should be less than 0 which is not possible because of the square term.
- Therefore, the function is nowhere concave down.
- Also, there is no inflection point and concavity does not change.

### Problem 4

How can we calculate the concave nature of a function if $f\text{'}\text{'}\left(x\right)$is never unknown and is never equal to zero?

Answer: For all x in any interval, $f\text{'}\text{'}\left(x\right)$must be greater than or equal to 0 (i.e. non-negative) in order for f(x) to be concave up.

Similarly $f\text{'}\text{'}\left(x\right)$must be non-positive in order for f(x) to be concave down.

Since $f\text{'}\text{'}\left(x\right)$may not be continuous and therefore shift polarity without crossing the x-axis, simply saying $f\text{'}\text{'}\left(x\right)$greater than zero 0 is insufficient to evaluate the concave nature of f(x).

### Problem 5

Is the product and sum of two concave functions concave as well?

Answer: Consider two functions, f and g, with $f\text{'}\text{'}\left(x\right)$>0 and $g\text{'}\text{'}\left(x\right)$>0, respectively.

- The second derivative of f+ g is f"+g", which is a positive number.
- As a result, their sum is concave up.
- The derivative f"g+2f'g'+fg", is obtained by taking the second derivative of fg.
- The words f" and g" are both positive, but the other terms can have any sign, so the expression as a whole may not be necessarily positive.

## Formulas

Let a function g be continuous on closed interval $\left[c,d\right]$ and twice differentiable on open interval $\left(c,d\right)$.

- The first derivative test tells us that:

(i) If the function's first derivative is greater than zero, that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$, then the function g is increasing on interval $\left[c,d\right]$.

(ii) If the function's first derivative is less than zero, that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$, then the function g is decreasing on interval $\left[c,d\right]$.

- The second derivative test, tells us that:

(i) If the second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$, then the function g is concave up on interval $\left[c,d\right]$.

(ii) If the second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$, then the function g is concave down on interval $\left[c,d\right]$.

- The power rule of differentiation is: $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$, where n is any real number.
- The sum/difference rule of differentiation is: $\frac{d}{dx}\left(f\pm g\right)=\frac{d}{dx}\left(f\right)\pm \frac{d}{dx}\left(g\right)$, where f and g are differentiable functions in x.

## Common Mistakes

- Students often make errors in determining the form of a concave graph. It is normal to make the mistake of identifying the interval where the graph is concave up or concave down.
- Students should use the second derivative test to define a function's interval to avoid making an error in determining the concave nature of the function and how concavity changes.

## Context and Applications

This topic is significant in the professional exams for undergraduate and postgraduate courses, especially for:

- B.Sc. in Science
- M.Sc. in Science
- Bachelor of Mathematics

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