1. If f(x0 > 0 on an interval, then f is increasing on that interval.
2. If f(x0 < 0 on an interval, then f is decreasing on that interval.
Example:
Find where the function f(x) = x^{3} – 12x + 1 is increasing and where it is decreasing.
Solution:
Step 1: Find the derivative of f
f ‘(x) = 3x^{2} – 12 = 3(x^{2} – 4) = 3(x –2) (x + 2)
Step 2: Set f ‘(x) = 0 to get the critical numbers
f ‘(x) = 3(x –2) (x + 2) = 0
x = 2, –2
Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f ‘(x) for each of the intervals. Use the increasing/decreasing test to determine whether f(x) is increasing or decreasing for each interval.
Interval |
x – 2 |
x + 2 |
f ‘(x) |
f |
x < – 2 |
– |
– |
+ |
Increasing on |
–2 < x < 2 |
– |
+ |
– |
Decreasing on (–2, 2) |
x > 2 |
+ |
+ |
+ |
Increasing on |
Suppose that c is a critical number of a continuous function f.
1. If f ‘ changes from positive to negative at c, then f has a local maximum at c.
2. If f ‘ changes from negative to positive at c, then f has a local minimum at c.
3. If f ‘ does not change sign at c (f ‘ is positive at both sides of c or f ‘ is negative on both sides), then f has no local maximum or minimum at c.
Example:
Find the local maximum and minimum values of the function f(x) = x^{4} – 2x^{2} + 3
Solution:
Step 1: Find the derivative of f
f ‘(x) = 4x^{3} – 4x = 4x(x^{2} –1) = 4x(x –1)(x +1)
Step 2: Set f ‘(x) = 0 to get the critical numbers
f ‘(x) = 4x(x –1)(x +1) = 0
x = –1, 0, 1
Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f ‘(x) for each of the intervals.
Interval |
4x |
x –1 |
x + 1 |
f ‘(x) |
x < –1 |
– |
– |
– |
– |
–1 < x < 0 |
– |
– |
+ |
+ |
0 < x < 1 |
+ |
– |
+ |
– |
x > 1 |
+ |
+ |
+ |
+ |
Step 4: Use the first derivative test to find the local maximum and minimum values.
f ‘(x) goes from negative to positive at x = –1, the First Derivative Test tells us that there is a local minimum at x = –1.
f (–1) = 2 is the local minimum value.
f ‘(x) goes from positive to negative at x = 0, the First Derivative Test tells us that there is a local maximum at x = 0.
f (0) = 3 is the local maximum value.
f ‘(x) goes from negative to positive at x = 1, the First Derivative Test tells us that there is a local minimum at x = 1.
f (1) = 2 is the local minimum value.
We can also use the Second Derivative Test to determine maximum or minimum values.
The Second Derivative Test
Suppose f ‘’ is continuous near c,
1. If f ‘(c) = 0 and f ‘’(c) > 0, then f has a local minimum at c.
2. If f ‘(c) = 0 and f ‘’(c) < 0, then f has a local maximum at c.
Example:
Use the Second Derivative Test to find the local maximum and minimum values of the function f(x) = x^{4} – 2x^{2} + 3
Solution:
Step 1: Find the derivative of f
f ‘(x) = 4x^{3} – 4x = 4x(x^{2} –1) = 4x(x –1)(x +1)
Step 2: Set f ‘(x) = 0 to get the critical numbers
f ‘(x) = 4x(x –1)(x +1) = 0
x = –1, 0, 1
Step 3: Find the second derivative
f ‘’(x) = 12x^{2} – 4
Step 4: Evaluate f ‘’at the critical numbers
f ‘’(–1) = 8 > 0, so f (–1) = 2 is the local minimum value.
f ‘’(0) = – 4 < 0, so f (0) = 2 is the local maximum value.
f ‘’(1) = 8 > 0, so f (1) = 2 is the local minimum value.
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