# Maxima and Minima using Derivatives

A maximum is a high point and a minimum is a low point.

The plural of Maximum is

**Maxima**The plural of Minimum is

**Minima**Maxima and Minima are collectively called

**Extrema**

### Local Maximum or Local Minimum

**First** we need to choose an interval:

The local **maximum** is the point where the height of the function at "a" is greater than (or equal to) the height anywhere else in that interval.

**Local Maximum**: $f(a) ≥ f(x)$ for all x in the interval
**Local Minimum**: $f(a) ≤ f(x)$ for all x in the interval

### Global (or Absolute) Maximum and Minimum

The maximum or minimum over the **entire function** is called an "Absolute" or "Global" maximum or minimum. There is **ONLY **one **global maximum **and one **global minimum**.

## Finding Maxima and Minima using Derivatives

In a smoothly changing function a **maximum **or **minimum **is always where the function **flattens out** except for a **saddle point (**A stationary point but not an extremum).

**Note: **When a function **flattens out**, the **slope **is **zero**. **Derivatives **is helpful to find where the **slope **is **zero**.

**Steps to Find the maxima and minima for**: $y = 5x^3 + 2x^2 − 3x$

**Steps to Find the maxima and minima for**: $y = 5x^3 + 2x^2 − 3x$

Given a function $y = f(x) = 5x^3 + 2x^2 − 3x$

Find slope of $f(x)$ using derivatives $\frac{df}{dx}$

$f'(x) = \frac{df}{dx} = 15x^2 + 4x − 3$ .....(i)

Take a

**second derivative**of the first slope $\frac{d}{dx}(\frac{df}{dx})$ Note: First slope can be found from Step 2 $f''(x) = \frac{d( 15x^2 + 4x − 3 )}{dx} = 30x+4$If the

**second derivative is:**less than 0, it is a local maximum

greater than 0, it is a local minimum

equal to 0, then the test fails (Could be a saddle point)

**At**$x = -\frac{3}{5}$**At**$x = \frac{1}{3}$ $f''(x) = 30(\frac{1}{3}) + 4 = 14$ which is greater than 0, so local maximum is at $x = \frac{1}{3}$

## Concavity

Taking the derivative f'(x) of a function f(x) gives the slope.

When the slope continually

**increases**, the function is**concave upward**.When the slope continually

**decreases**, the function is**concave downward**.

Using the **second derivative** f''(x):

When the second derivative is

**positive**, the function is**concave upward**.When the second derivative is

**negative**, the function is**concave downward**.

Link: - MathsIsFun: Maxima and Minima of Functions - MathsIsFun: Finding Maxima and Minima using Derivatives - KhanAcademy: Maxima, Minima and Saddle Points - KhanAcademy: Concavity

Last updated