MCV4U – Prepared by Mrs. Snyder
SUMMARY OF DERIVATIVE TESTS
Definition: A critical point or critical number of a function f is a point 𝑥 = 𝑐 in the domain of f
such that either 𝑓 (𝑐) = 0 or 𝑓 (𝑐) does not exist. (Be careful to rule out any point that is not
within the domain of the function. For example, any zero of the denominator of a rational
function is never a critical number, since it is not in the domain of the function.)
The Closed Interval Test for Absolute Maximum/Minimum
Purpose: To find the absolute maximum and minimum points/values of a continuous function f
on a closed interval:
1. Find all critical points of f in the given interval. Ignore any critical point that is not in the
interval.
2. Evaluate f at the critical point(s) found in step 1, as well as at the two endpoints of the
interval.
3. The point(s) of the largest value of f is the absolute maximum(s), the point(s) of the
smallest value is the absolute minimum(s).
Increasing and Decreasing Interval Test
Purpose: To determine is f(x) is increasing or decreasing on an interval.
1. Find the critical points of f.
2. Use the critical point(s) and any vertical asymptotes to create an
interval chart.
3. Determine the sign of 𝑓 (𝑥) at one value in the intervals.
4. If 𝑓 (𝑥) > 0, then f(x) is increasing on the interval (a,b).
5. If 𝑓 (𝑥) < 0, then f(x) is decreasing on the interval (a,b).
6. If 𝑓 (𝑥) = 0, then f(x) is constant on the interval (a,b).
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MCV4U – Prepared by Mrs. Snyder
The First Derivative Test for Local Maximum/Minimum
Purpose: To find the local maximum and minimum points of a continuous function f within its
domain.
1. Find all critical points and vertical asymptotes of f.
2. Use the critical point(s) and any vertical asymptotes to create an interval
chart.
3. Determine the sign of 𝑓 (𝑥) at one value in the intervals.
4. For each critical point c (only; but not for a vertical asymptote, which
cannot be an extreme point):
NOTE: While a vertical
asymptote will never be the
location of a local extreme
point, it nevertheless needs
to be included in the
process because it
separates the curve into 2
distinct parts which might
behave very differently
from each other.
a) If 𝑓 changes from positive to negative at c, then f has a local maximum at c.
b) If 𝑓 changes from negative to positive at c, then f has a local minimum at c.
c) If 𝑓 does not change sign at c, then f has neither a local maximum nor a
minimum at c.
The Second Derivative Test for Local Maximum/Minimum
Purpose: To find the local maximum and minimum points of a differentiable function f within its
domain. Note: It only works at a critical point where 𝑓 ′(𝑐) exists and is nonzero.
1. Find all critical points of f.
2. At each critical point c of f, evaluate𝑓 ′(𝑐).
a. If 𝑓 (𝑐) = 0 and 𝑓 ′(𝑐) > 0, then f has a local minimum at c.
b. If 𝑓 (𝑐) = 0 and 𝑓 ′(𝑐) < 0, then f has a local maximum at c.
The test is inconclusive if 𝑓 (𝑐) doesn’t exist, or 𝑓 ′(𝑐) = 0, or 𝑓 ′(𝑐) doesn’t exist. The
First Derivative Test must be used in such cases.
Comment: This test is usually easier to perform than the First Derivative Test (assuming that 𝑓
is easy to find) − there is only one calculation to do for/at each critical point; there is no need to
check the intervals on either side, nor do you need to worry about any vertical asymptote. But
unfortunately, this test doesn’t always work.
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MCV4U – Prepared by Mrs. Snyder
The Test for Concavity and Inflection Points
Purpose: To find the concavity and inflection points of a continuous function f within its domain.
1. Find all points c such that 𝑓 ′(𝑐) = 0, or 𝑓 ′(𝑐) doesn’t exist; and all vertical asymptotes
of f.
2. Separate the x-axis into one or more intervals using the candidate points found in step 1
(and vertical asymptotes, if any, of the function).
Note: Again, a vertical asymptote will never be the location of an inflection point. But it
needs to be included in the process because it separates the curve into 2 distinct parts
which might have different concavities across the asymptote.
3. Within each interval thus obtained, pick any convenient point and
evaluate 𝑓 ′(𝑥) at the point. Do this for every interval. For each interval,
a) If 𝑓 (𝑥) > 0, then f(x) is concave up on the interval.
b) If 𝑓 (𝑥) < 0, then f(x) is concave down on the interval.
For each candidate point c (only; but not for a vertical asymptote, which
cannot be an inflection point):
c) If 𝑓 ′ changes sign (either from positive on the left of c to negative on the right of
c, or from negative to positive) at c, then f has an inflection point at c.
d) If 𝑓 ′ does not change sign at c (i.e., 𝑓 ′ remain either positive or negative on both
sides of c), then f does not have an inflection point at c.
Comment: The steps are virtually identical to those of the First Derivative Test for Local
Maximum/Minimum, except all the calculations are done with the second derivative of f instead.
The First Derivative Test for Absolute Maximum/Minimum
Purpose: To find the absolute maximum or minimum point of a continuous function f within an
interval containing only one critical point.
1. Find the critical point 𝑥 = 𝑐 of f on the interval.
2. Choose any point within the interval on the left of c and another point within the interval
on the right of c. Evaluate 𝑓 (𝑥) at those two chosen points:
a) If 𝑓 changes from positive on the left of c to negative on the right of c, then f has an
absolute maximum at c.
b) If 𝑓 changes from negative to positive at c, then f has an absolute minimum at c.
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MCV4U – Prepared by Mrs. Snyder
CHART SUMMARY OF DERIVATIVE TESTS
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