Local Maximum/Minimum of Continuous Functions

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The intervals where a continuous function f (x) increases are exactly the intervals where f ’(x) > 0.

The intervals where a continuous function decreases correspond to the intervals where f ’(x) < 0.

KNOWN FACTS

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a. Identify all critical pointsb. Look at the sign of f’(x) before and after a “turn”Look at the sign of f’(x) before and after a “plateau”

From A Graph

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Being a critical point does not guarantee that there is a change from increasing to decreasing at that point.

A point where a continuous function changes from increasing to decreasing, or from decreasing to increasing is a critical point.

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A point (c, f(c)) where a function changes from increasing to decreasing is called a LOCAL

MAXIMUM point of the function. The value f(c) is a local maximum value of the function.

A point (c , f(c)) where a function changes from decreasing to increasing is called a LOCAL

MINIMUM point of the function. The value f(c) is a local minimum value of the function.

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The intervals where a continuous function f(x) increases are the intervals where f’(x)>0.

The intervals where a continuous function decreases are the intervals where f’(x)<0.

In addition it is known that a local maximum the function changes from increasing to decreasing, and at a local minimum the function changes from decreasing to increasing.

REMEMBER

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THE FIRST DERIVATIVE TEST FOR LOCAL MAXIMA AND MINIMA

Let p be a critical point of a continuous function f.

• If f ’ changes from negative - for values less than p- to positive -for values greater than p-, f has a local minimum at p and its value is f(p). This is equivalent to saying that the function changes from decreasing to increasing at the point (p , f(p)).

• If f ‘changes from positive -for values less than p- to negative -for values greater than p-, f has a local maximum at p, and its value is f(p). This is equivalent to saying that the function changes from increasing to decreasing at the point (p , f(p)).

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STEPS TO APPLY THE FIRST DERIVATIVE TEST

To determine algebraically the intervals where a function increases or decreases these are the steps to follow.

1. Find the domain of the function. 2. Find the critical points of the function. 3. On a number line label the domain and all the critical points of the function. Notice that on each of the intervals produced after these points are determined, the derivative is always either positive or negative. 4. From each of the intervals found in step (3) choose a point, test point.

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5. Determine the sign of the derivative at each of the test points. The sign of the derivative at a test point gives the sign of the derivative on the interval it was chosen from. The sign of the derivative determines whether the function is increasing or decreasing on the interval.

On the intervals where the derivative is positive the function is increasing and on the intervals where the derivative is negative the function is decreasing.

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Example 1Consider the function y=2x-1 with derivative y ’(x)=2Follow the steps described above to find the intervals where the function is increasing or decreasing. Use that information to determine any local max/min points.

I. The domain of this function is all the real number

II. The critical points are the solution of y’(x)=0, or y’(x) undefined. Since y’(x)=2 neither of these equations has a solution. Hence there are no critical points.

III.

IV. Test Points 1

V. Sign of y ‘ (x) +

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Notice that the second derivative test is not necessary since there are no critical points.

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EXAMPLE 2 Consider the function f(x)=2x2-x+3 and its derivative f’(x)=4x-1

I. The domain is all the real numbers or (-∞,∞)II. The critical points are the solution to f’(x)=0, or f’(x) undefined.

f’(x)=0 4x-1=0 or x=1/4 f’(x) undefined has no solution. III.

IV. -1 1V. Sign of f’ −− +

The function is decreasing on (-∞,1/4) and increasing on (1/4,∞). There is a local minimum at x=1/4 and its value is f(1/4)=2.875.

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To use the second derivative test we need f "(x)=4. Since f" is positive for any value of x, it is positive at the critical point x=0.25. Hence the function is concave up and the point becomes a local minimum point on the graph, or f(0.25) a local minimum value of the function.

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EXAMPLE 3Consider the function with derivative I. Domain. Solve the inequality

We find it by solving (x-1)(2-x)=0. The solutions are x=1,-2. Now use the number line to find the solution to the inequality.

0.5 1.5 2.5 Sign of (x-1)(2-x) __ + _

The solution to the inequality is [1,2], which is the domain of the function.

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II. Find the critical points. Need to solve h’(x)=0 or h’(x) undefined.

Notice that the end points of the domain are automatically critical points.

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III.

IV. Test Points 1.1 1.7

V. Sign of h’ + __

So the function increases on [1,1.5] and decreases on [1.5,3].

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THE SECOND DERIVATIVE TEST FOR LOCAL MAXIMA AND MINIMA

Let p be a critical point of a continuous function f

If the function is concave up at the critical point, it is a local minimum. It means that if p is a critical point and f "(p) > 0, the point (p , f(p)) is a local minimum of the function f

If the function is concave down at the critical point, it is a local maximum. It means that if p is a critical point and f "(p)< 0, the point (p , f(p)) is a local maximum of the function f

Sometimes the function is so flat that f "(p)= 0, the (p , f(p)) could be a local maximum, minimum, or neither. In this case go back to the first derivative test since one can not conclude anything from the second derivative

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Second Derivative Inconclusive

Local Minimum

at x=0

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Second Derivative Inconclusive

Local Maximum

at x=0

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Second Derivative InconclusiveNeither a

Local Max/Min at

x=0

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EXERCISE Let

a. Find its largest domain

b. Find its critical points and use the first derivative test to determine where the function is increasing or decreasing

c. Use the second derivative test, if possible to corroborate your answer from (b).

d. Determine the behavior of the function toward the end points of the domain.

e. Use technology to graph the function, not its derivative, and double check the answers you gave in (a)-(c).