# How to Find Maxima and Minima of a Function?

Maxima and minima are the maximum or the minimum value of a function in a given range. In this step-by-step guide, you learn how to find the maxima and minima of a function. Maxima and minima are known as the extrema of a function. Maxima will be the highest point of the curve in the given range and the minimum will be the lowest point of the curve.

## Step by Step guide to find maxima and minima of a function

The maximum and minimum are peaks and valleys in the curve of a function. For a function, there can be any number of maximum or minimum. In calculus, we can find the maximum and minimum values of each function without even looking at the function diagram. Maxima will be the highest point of the curve in the given range and the minimum will be the lowest point of the curve.

The combination of maximum and minimum is extrema. In the picture below, we see different peaks and valleys in the diagram. At $$x=a$$ and at $$x=0$$, we get maximum values of the function, and at $$x=b$$ and $$x=c$$, we get minimum values of the function. All the peaks are the maxima and the valleys are the minima.

There are two types of maximum and minimum in a function, which are:

• Local maxima and minima
• Absolute or global maxima and minima

### Local maxima and minima

Local maximum and minimum are the maximum and minimum of a function that is generated in a certain interval. The local maximum is the value of a function at a point in a given interval where the values of the function close to that point are always less than the value of the function at that point. While the local minimum is the value of the function at a point where the values of the function close to that point are greater than the value of the function at that point.

• Local maxima: A point $$x=b$$ is a point of local maximum for $$f(x)$$ if in the neighborhood of $$b$$ i.e in $$(b−𝛿, b+𝛿)$$ where $$𝛿$$ can be made arbitrarily small, $$f(x)< f(b)$$ for all $$x ∈ \frac{(b−𝛿, b+𝛿)}{{b}}$$. This simply means that if we consider a small region (interval) around $$x=b, f(b)$$ should be the maximum in that interval.
• Local minima: A point $$x=a$$ is a point of local minimum for $$f(x)$$ if in the neighbourhood of $$a$$, i.e. in $$(a−𝛿,a+𝛿)$$, (where $$𝛿$$ can have arbitrarily small values), $$f(x)>f(a)$$ for all $$x ∈ \frac{(a−𝛿,a+𝛿)}{{a}}$$. This means that if we consider a small interval around $$x=a$$, $$f(a)$$ should be the minimum in that interval.

In the image given below, we can see that $$x = b$$ and $$x = d$$, are the local maxima, and $$x=a$$ and $$x = c$$, are the local minima.

### Absolute maxima and minima

The highest point of a function in the whole domain is known as the absolute maximum of the function while the lowest point of the function within the entire domain of the function, is known as the absolute minimum of the function. There can only be one absolute maximum of a function and one absolute minimum of the function over the entire domain. The absolute maxima and minima of the function can also be called the global maxima and global minima of the function.

• Absolute maxima: A point $$x=a$$ is a point of global maximum for $$f(x)$$ if $$f(x)≤ f(a)$$ for all $$x∈ D$$ (the domain of $$f(x)$$).
• Absolute minima: A point $$x=a$$ is a point of global minimum for $$f(x)$$ if $$f(x) ≥ f(a)$$ for all $$x∈ D$$ (the domain of $$f(x)$$).

In the image given below, point $$x=a$$ is the absolute maxima of the function and at $$x=b$$ is the absolute minima of the function.

### Finding maxima and minima of a function

The maximum and minima of a function can be calculated using the first-order derivative test and the second-order derivative test.

First-order derivative test for maxima and minima

The first derivative of the function shows the slope of the function. Near a maximum point, the slope of the curve increases with going to the maximum point, then decreases to zero at the maximum point, and then decreases as we move away from the maximum point. Similarly, near the minimum point, the slope of the function decreases as we move toward the minimum point, then becomes 0 at the minimum point, and then increases as we move away from the minimum point.

Suppose we have a function $$f$$ that is continuous at the critical point and is defined in the open interval $$I$$ and $$f’(c)= 0$$ (slope is $$0$$ at $$c$$). Then we examine the value $$f'(x)$$ at the left point of the curve and the right of the curve and examine the nature of $$f'(x)$$, then we can say that the given point will be:

• Local maxima: If $$f’(x)$$ changes the sign from positive to negative as $$x$$ increases via point $$c$$, then $$f(c)$$ gives the maximum value of the function in that range.
• Local minima: If $$f’(x)$$ changes the sign from negative to positive as $$x$$ increases via point $$c$$, then $$f(c)$$ gives the minimum value of the function in that range.
• Point of inflection: If the sign of $$f’(x)$$ doesn’t change as $$x$$ increases via $$c$$, and the point $$c$$ is neither the maxima nor minima of the function, then the point $$c$$ is called the point of inflection.

Second-order derivative test for maxima and minima

In the second-order derivative test for maxima and minima, we find the first derivative of the function, and if it gives the value of the slope equal to $$0$$ at the critical point $$x=c (f’(c)= 0)$$, then we find the second derivative of the function.

If the second derivative of the function exists in the given range, the given point will be as follows:

• Local maxima: If $$f”(c)<0$$
• Local minima: If $$f”(c)>0$$
• Test fails: If $$f”(c)=0$$

### Maxima and Minima of a Function – Example 1:

Find the maximum and minimum value of the function. $$2x^3-3x^2+6$$

Solution:

Using a second-order derivative test for the maxima and minima of a function:

Taking the first-order derivative of:

$$y=2x^3 – 3x^2+ 6$$, $$(eq\ 1)$$

Differentiate both of sides (eq $$1$$):

$$\frac{dy}{dx}= \frac{d(2x^3)}{dx}- \frac{d(3x^2)}{dx} + \frac{d(6)}{dx}$$

⇒ $$\frac{dy}{dx}=6x^2- 6x + 0$$

⇒ $$\frac{dy}{dx }= 6x^2- 6x$$ , $$(eq\ 2)$$

Putting $$\frac{dy}{dx}= 0$$ to find critical points,

⇒$$6x^2- 6x = 0$$

⇒ $$6x(x -1)= 0$$

⇒ $$x= 0,1$$

The critical points are $$(0,1)$$.

Differentiate both of sides of (eq $$2$$):

⇒ $$\frac{d^2y}{dx^2 }= \frac{d(6x^2)}{dx} – \frac{d(6x)}{dx}$$

⇒ $$\frac{d^2y}{dx^2}= 12x – 6$$

Now, put the values of $$x$$ and find the max or min value.

At $$x = 0$$, $$\frac{d^2y}{dx^2}= 12(0)-6=-6<0$$, hence $$x = 0$$ is a point of maxima.

At $$x = 1$$, $$\frac{d^2y}{dx^2}= 12(1)-6= 6 > 0$$, hence $$x=1$$ is a point of minima

## Exercises for Maxima and Minima of a Function

### Find the extrema of each function.

1. $$\color{blue}{f(x)=x^3-3x^2-9x+12}$$
2. $$\color{blue}{f\left(x\right)=4sin^2x+1,\:0\le \:x\le 8}$$
3. $$\color{blue}{f\left(x\right)=x^2,\:0\le \:x\le 3}$$
4. $$\color{blue}{f\left(x\right)=2x^2-2x^3}$$
1. $$\color{blue}{ max:(-1, 17), min:(3,-15)}$$
2. $$\color{blue}{max:(\frac{\pi }{2}, 5), min:(0,1)}$$
3. $$\color{blue}{ max:(3, 9), min:(0,0)}$$
4. $$\color{blue}{ max:(\frac{2}{3},\frac{8}{27}), min:(0,0)}$$

### What people say about "How to Find Maxima and Minima of a Function?"?

No one replied yet.

X
30% OFF

Limited time only!

Save Over 30%

SAVE $5 It was$16.99 now it is \$11.99