Introduction to Derivatives

How do you find intervals of increase and decrease and points of extremum?

Q: What is the difference between a critical value and an extremum?

A critical value is an input x where the derivative equals zero or does not exist. An extremum is the actual peak or valley point on the curve, which occurs only if the derivative changes sign across that critical value.

Q: Can a function have a horizontal tangent without an extremum?

Yes. A classic example is f(x) = x^3. At x = 0, the derivative is zero, creating a horizontal tangent line. However, the derivative is positive on both sides of zero, meaning the curve continues rising rather than turning around.

Q: How do you find the y-coordinate of a local extremum?

Substitute the critical x-value back into the original function f(x), never into f'(x). The derivative only confirms whether the point is a maximum or minimum, whereas the original equation gives its height.

Q: Why do we test values instead of just graphing the function?

Testing values using calculus provides algebraic certainty. Graphs can hide tiny fluctuations or deceive your eyes around closely spaced roots and subtle turns, whereas sign testing pinpoints exact turning points every time.

Q: What tools are available if I get stuck solving a problem?

You can ask Lessi for step-by-step help. Lessi is an AI teacher, and the first 60 minutes of lessons are free with an account. The Ask Lessi feature answers individual questions free from those same minutes.

Quick answer

To find intervals of increase and decrease and points of extremum, calculate the first derivative f'(x), set it equal to zero, and solve for candidate values. Test the sign of f'(x) across each interval. Where f'(x) is positive, the function increases; where negative, it decreases. A switch from positive to negative marks a local maximum, while negative to positive marks a local minimum.

The topic explained

The three-step process for curve analysis

To find intervals of increase and decrease, begin by differentiating the function f(x) to obtain f'(x). Next, set f'(x) = 0 and solve for x. These solutions, along with any points where the derivative does not exist, are your critical points. They represent the only x-values where the graph can change its direction.

Split the real number line into test intervals bounded by these critical values. Pick a single test number inside each interval and plug it into f'(x). If f'(x) > 0, the function rises throughout that entire interval. If f'(x) < 0, the function falls.

Reading a sign table and classifying extrema

Organize your findings in a sign table with rows for the test intervals, the sign of f'(x), and the behavior of f(x). A zero derivative alone does not guarantee a peak or a valley; the sign must actually change across the critical point.

If f'(x) changes from positive to negative, the function climbs and then drops, creating a local maximum. If f'(x) changes from negative to positive, the curve bottoms out at a local minimum. If the sign does not switch, such as with f(x) = x^3 at x = 0, the point is neither a maximum nor a minimum.

A concrete quadratic example

Take the function f(x) = x^2 - 6x. The derivative is f'(x) = 2x - 6. Setting 2x - 6 = 0 yields x = 3 as the only critical number. Test the left interval with x = 0 to get f'(0) = -6 (negative). Test the right interval with x = 4 to get f'(4) = 2 (positive).

The function decreases on (-infinity, 3) and increases on (3, infinity). Because the derivative shifts from negative to positive at x = 3, there is a local minimum. Substitute x = 3 back into the original function: f(3) = 3^2 - 6(3) = -9. The local minimum point is (3, -9).

Common errors to avoid

Always test values inside the derivative f'(x), not the original function f(x). Substituting a test number into f(x) reveals height, not direction.

Finally, distinguish between the location of an extremum and the extremum itself. An x-value tells where the peak or valley sits, while the y-value gives the extreme value. A full point of extremum always requires both coordinates: (x, y).

Free printable worksheet

Calculus Practice: Intervals of Increase, Decrease, and Extrema

Designed for 12th-grade calculus students. Differentiate each function, find all critical points, determine the intervals of increase and decrease, and state any local extrema as ordered pairs (x, y).

Print the worksheet Regenerate new worksheet

Free, no sign-up. The answer key prints on the second page. Regenerating a new worksheet on the same topic needs a free account.

Personal AI lessons and practice, from preschool to university

Free printable worksheetlessoons.com

Calculus Practice: Intervals of Increase, Decrease, and ExtremaName:

Find the critical points of f(x) = x^2 - 8x + 12.
Determine the intervals of increase and decrease for f(x) = x^2 - 8x + 12.
Identify and classify the extremum of f(x) = x^2 - 8x + 12 as a coordinate pair.
Find the critical points and intervals of increase and decrease for f(x) = -x^2 + 10x - 7.
Find the local extrema of f(x) = -x^2 + 10x - 7.
Find the intervals of increase and decrease for f(x) = 2x^3 - 6x.
Classify all local extrema for f(x) = 2x^3 - 6x.
Analyze f(x) = x^3 - 3x^2 + 3x - 1 for local extrema. Does x = 1 produce an extremum?

Common questions

What is the difference between a critical value and an extremum?

A critical value is an input x where the derivative equals zero or does not exist. An extremum is the actual peak or valley point on the curve, which occurs only if the derivative changes sign across that critical value.

Can a function have a horizontal tangent without an extremum?

Yes. A classic example is f(x) = x^3. At x = 0, the derivative is zero, creating a horizontal tangent line. However, the derivative is positive on both sides of zero, meaning the curve continues rising rather than turning around.

How do you find the y-coordinate of a local extremum?

Substitute the critical x-value back into the original function f(x), never into f'(x). The derivative only confirms whether the point is a maximum or minimum, whereas the original equation gives its height.

Why do we test values instead of just graphing the function?

Testing values using calculus provides algebraic certainty. Graphs can hide tiny fluctuations or deceive your eyes around closely spaced roots and subtle turns, whereas sign testing pinpoints exact turning points every time.

What tools are available if I get stuck solving a problem?

You can ask Lessi for step-by-step help. Lessi is an AI teacher, and the first 60 minutes of lessons are free with an account. The Ask Lessi feature answers individual questions free from those same minutes.

Still have a question? Ask Lessi

Any question about this subject. Lessi answers right here, step by step.

Try asking

Asking needs a free account; your question is kept while you sign up. Answers are written by AI and can contain mistakes. Lessi doesn’t give medical or legal advice.