Skip to content
Study Hall
Lesson 1 of 4Curve Sketching
Unit 3 · Lesson 1 of 4

Increasing and Decreasing Functions and the First Derivative Test

The sign of the derivative tells you whether a graph is rising or falling. Where the derivative is zero or undefined you find the critical numbers, and the first derivative test tells you which of them are local maxima, local minima or neither.

6 min read10 practice questionsInteractiveAsk AI about this →

Key things to remember

  1. 1If on an interval, is increasing there; if , is decreasing there.
  2. 2A critical number is a in the domain of with or undefined.
  3. 3First derivative test: changes to at gives a local maximum; to gives a local minimum; no sign change gives neither.
  4. 4 alone does not guarantee an extremum: has but no maximum or minimum at .
  5. 5Find the height of an extremum from , not .

Picture a hiking trail drawn as a graph of height against distance. Uphill stretches, downhill stretches, the summits and the valley floors are the first things a hiker wants to know, and they are exactly the features you want when you sketch a function. A graphing calculator can show you a picture, but it cannot promise that the window shows every peak and valley. The derivative can. Because is the slope of the tangent at every point, its SIGN tells you which way the graph is heading, and the places where the slope is zero (or undefined) are the only places where a continuous graph can turn around.

A function is INCREASING on an interval if its values go up as moves to the right: whenever in the interval, . It is DECREASING if its values go down: whenever , . On a smooth graph, a rising curve has tangents that slope upward, so a positive derivative means increasing, and a negative derivative means decreasing.

On an intervalTangent linesThe function
slope upwardincreasing (graph rises left to right)
slope downwarddecreasing (graph falls left to right)
at every pointhorizontalconstant (a horizontal line)
The sign of the first derivative and what the graph does.
f(x) = x3 − 3x, f′(x) = 3x2 − 3 = 3(x + 1)(x − 1)xy−2−112−22local max (−1, 2)local min (1, −2)xx < −1−1−1 < x < 11x > 1f′(x)+0−0+f(x)increasing ↗maxdecreasing ↘minincreasing ↗First derivative test: f′ changes from + to − at x = −1, so (−1, 2) is a local maximum;f′ changes from − to + at x = 1, so (1, −2) is a local minimum.
Fig 1The graph of f(x) = x³ − 3x with a local maximum at (−1, 2) and a local minimum at (1, −2), and the sign chart of f′(x) = 3x² − 3 underneath: positive, then negative, then positive.

Read the figure from left to right. For the derivative is positive, so the graph climbs. At the slope is and the graph levels off at the top of a hill. Between and the derivative is negative and the graph falls. At the slope is again, at the bottom of a valley, and after that the graph climbs forever. The function changes direction only at the two places where .

VideoEddie Woo explains what the sign of the first derivative tells you: positive means the curve is rising, negative means it is falling.Watch on YouTube

Critical numbers

Critical number
A CRITICAL NUMBER of is a value in the domain of where or does not exist. The point is called a critical point. On a graph, is a horizontal tangent, and not existing is usually a sharp corner, a cusp or a vertical tangent.

Why care about critical numbers? A smooth function cannot switch from rising to falling without passing through a slope of , and a function with a corner can switch at the corner. So every LOCAL MAXIMUM (a point higher than all nearby points) and every LOCAL MINIMUM (a point lower than all nearby points) inside an open interval happens at a critical number. For polynomials exists everywhere, so the critical numbers are just the solutions of . For a function such as , the derivative does not exist at , so is a critical number even though is never zero: the graph has a cusp there and a local minimum at .

The first derivative test

  1. Find and factor it as far as you can.
  2. Find the critical numbers: solve , and note any in the domain where does not exist.
  3. Mark the critical numbers on a number line. They split it into intervals.
  4. Pick a test value in each interval and find the sign of there (the signs of the factors are enough).
  5. Classify each critical number: changes from to gives a LOCAL MAXIMUM; from to gives a LOCAL MINIMUM; no sign change gives neither.
  6. Substitute each critical number into (not ) to find the height of the maximum or minimum.
A zero slope is not always a turning point
For , and , so is a critical number. But on both sides of , so the graph rises, flattens for an instant, and keeps rising. There is no maximum or minimum at . That is why the test looks at the sign on BOTH sides, not just at the zero.

Two of the curriculum's sample functions show the simplest cases. For , is positive everywhere, so the line increases on all of and has no critical numbers and no extrema. For is zero at , negative before it and positive after it. So decreases on , increases on and has a local (in fact absolute) minimum at , the vertex of the parabola.

IntervalTest valueSign of is
decreasing
increasing
decreasing
increasing
Sign chart for f(x) = x⁴ + 4x³ − 3x² − 18x, whose derivative factors as 2(x + 3)(2x² − 3). The critical numbers are −3 and ±√6/2 (about ±1.22).

Reading the chart: goes to at , so there is a local minimum (the graph touches the -axis there). It goes to at , a local maximum of , and to at , a local minimum of . The lower of the two minimums is also the absolute minimum of the whole graph.

Loading interactive…
Drag a point along a polynomial and watch the tangent slope; the derivative is plotted point by point, positive where the graph rises and zero at each local maximum and minimum.
VideoA worked walkthrough of finding critical points and classifying them as local maximums or minimums with a sign chart of the first derivative.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1
For , find the intervals of increase and decrease and the local extrema.
Example 2
For , find the critical numbers exactly and classify them.
Example 3
Find and classify the critical numbers of .

Where marks get lost

  • Classifying every solution of as a maximum or minimum without checking the sign on both sides. In , is neither.
  • Substituting critical numbers into to find the height of a turning point. by definition; the height is .
  • Solving (the -intercepts) instead of (the horizontal tangents) when looking for turning points.
  • Forgetting the points where does not exist, such as the cusp of at .
  • Writing the intervals of increase as -values, or including a critical number in two intervals as if it were a whole interval. State intervals of , such as and .

Flashcards 8

Increasing on an interval
Decreasing on an interval
Critical number
First derivative test: to
+ 4 more

Practice

1
On the interval , . What does this tell you about ?
2
On which interval is decreasing?
3
has one critical number. What is it? Give a decimal.
4
Find the local maximum VALUE (the -coordinate) of .
5
How many critical numbers does have?
6
For , . What happens at ?
7
Why is a critical number of ?
8
Match each piece of information about with what it tells you about .
  1. A. has a local maximum at
  2. B. has a local minimum at
  3. C. is decreasing on that interval
  4. D. is increasing on that interval
on an interval
on an interval
changes from positive to negative at
changes from negative to positive at
9
Put the steps of the first derivative test in order.
  1. 1Substitute each extremum's -value into to find its height
  2. 2Use the sign changes of to classify each critical number
  3. 3Choose a test value in each interval between critical numbers
  4. 4Find and factor
  5. 5Solve and note where is undefined
10
What is the name for a number in the domain of where or does not exist? (Two words.)

Spot a mistake on this page? Let us know