If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 5

Lesson 4: Using the first derivative test to find relative (local) extrema

# Relative minima & maxima review

Review how we use differential calculus to find relative extremum (minimum and maximum) points.

## How do I find relative minimum & maximum points with differential calculus?

A relative maximum point is a point where the function changes direction from increasing to decreasing (making that point a "peak" in the graph).
Similarly, a relative minimum point is a point where the function changes direction from decreasing to increasing (making that point a "bottom" in the graph).
Supposing you already know how to find increasing & decreasing intervals of a function, finding relative extremum points involves one more step: finding the points where the function changes direction.

## Example

Let's find the relative extremum points of f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3, x, squared, minus, 9, x, plus, 7. First, we differentiate f:
f, prime, left parenthesis, x, right parenthesis, equals, 3, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis
Our critical points are x, equals, minus, 3 and x, equals, 1.
Let's evaluate f, prime at each interval to see if it's positive or negative on that interval.
Intervalx-valuef, prime, left parenthesis, x, right parenthesisVerdict
x, is less than, minus, 3x, equals, minus, 4f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
minus, 3, is less than, x, is less than, 1x, equals, 0f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0f is decreasing. \searrow
x, is greater than, 1x, equals, 2f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0f is increasing. \nearrow
Now let's look at the critical points:
xBeforeAfterVerdict
minus, 3\nearrow\searrowMaximum
1\searrow\nearrowMinimum
In conclusion, the function has a maximum point at x, equals, minus, 3 and a minimum point at x, equals, 1.