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: $y = 5x^3 + 2x^2 − 3x$

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

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

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

3. Slope is zero at extrema. So, use $\frac{df}{dx} = 0$ to solve for x. This would give the position of x-coordinates the local minima or maxima Solve: $15x^2 + 4x − 3 = 0$ The above equation is a quadratic ( $ax^2+bx+c=0$ ) $x = -\frac{3}{5}; x = \frac{1}{3}$
4. Take a second derivative of the first slope $\frac{d}{dx}(\frac{df}{dx})$ Note: First slope can be found from Step 2 $f''(x) = \frac{d( 15x^2 + 4x − 3 )}{dx} = 30x+4$

5. If the second derivative is:

   1. less than 0, it is a local maximum

   2. greater than 0, it is a local minimum

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

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