# Finding Maxima and Minima using Derivatives

Where will a function reach a high or low point? Calculus can help!

In a smoothly changing function a low point (a minimum) or high point (a maximum) occurs where the graph of the function **flattens out**:

(Not all flat points are maxima or minima, we could also have a **saddle point**)

It flattens out where the **slope is zero**.

So ... how do we find the slope? Using Derivatives! (Read that now if you don't already know.)

### Example: A ball is thrown in the air. Its height at any time t is given by:

h = 3 + 14t − 5t^{2}

### What is its maximum height?

Using derivatives we can find the slope of that function:

h = 0 + 14 − 5(2t)

= 14 − 10t

(See below this example for how we found that derivative.)

Now find when the **slope is zero**:

**1.4**

The slope will be zero at **t = 1.4 seconds**

And the height at that time will be

^{2}

**12.8**

And so:

The maximum height is **12.8 m** (at t = 1.4 s)

### A Quick Refresher on Derivatives

A derivative basically finds the slope of a function.

In the previous example we took this:

h = 3 + 14t − 5t^{2}

and came up with this derivative:

h = 0 + 14 − 5(2t)

= 14 − 10t

Which tells us the **slope** of the function at any time **t**

There are **rules** you can follow to find derivatives, and we used these rules:

- The slope of a
**constant**value (like 3) is 0 - The slope of the
**line**14t is 14 - A
**square**function like t^{2}has a slope of 2t, so 5t^{2}has a slope of 5(2t) = 10t - And then we added them up

Learn more at Derivative Rules

## How Do We Know it is a Maximum (or Minimum)?

We saw it on the graph! But otherwise ... derivatives come to the rescue again.

Take the **derivative of the slope** (the second derivative of the original function):

The Derivative of 14 − 10t is **−10**

This means the slope is continually getting smaller (−10): travelling from left to right the slope starts out positive (the function rises), goes through zero (the flat point), and then the slope becomes negative (the function falls):

A slope that gets smaller (and goes though 0) means a maximum.

This is called the **Second Derivative Test**

On the graph above I showed the slope before and after, but in practice we do the test **at the point where the slope is zero**:

### Second Derivative Test

When a function's **slope is zero at x**, and the **second derivative at x** is:

- less than 0, it is a local maximum
- greater than 0, it is a local minimum
- equal to 0, then the test fails (there may be other ways of finding out though)

"Second Derivative: less than 0 is a maximum, greater than 0 is a minimum"

### Example: Find the maxima and minima for:

y = 5x^{3} + 2x^{2} − 3x

The derivative (slope) is:

y = 15x^{2} + 4x − 3

Which is quadratic with zeros at:

- x = −3/5
- x = +1/3

Could they be maxima or minima? (Don't look at the graph yet!)

The second derivative is **y'' = 30x + 4**

At x = −3/5:

At x = +1/3:

(Now you can look at the graph.)

## Words

A high point is called a **maximum** (plural * maxima*).

A low point is called a **minimum** (plural * minima*).

The general word for maximum or minimum is **extremum** (plural **extrema**).

We say **local** maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby.

## One More Example

### Example: Find the maxima and minima for:

y = x^{3} − 6x^{2} + 12x − 5

The derivative (slope) is:

y = 3x^{2} − 12x + 12

Which is quadratic with only one zero at **x = 2**

Is it a maximum or minimum?

The second derivative is **y'' = 6x − 12**

At x = 2:

And here is why:

It is a **saddle point** ... the slope does become zero, but it is neither a maximum or minimum.

## Must Be Differentiable.

And there is an important technical point:

The function must be **differentiable** (the derivative must exist at each point in its domain).

### Example: How about the function f(x) = |x| (absolute value) ?

|x| looks like this: |

At x=0 it has a very pointy change!

In fact it is not differentiable there (as shown on the differentiable page).

So we can't use this method for the absolute value function.

(The function must also be continuous, but any function that is differentiable is also continuous.)