Calculus and Analytical Geometry

Lecture # 13

MTH 104

RELATIVE MAXIMA AND MINIMA

RELATIVE MAXIMA AND MINIMA

1. A function f is said to have a relative maximum at if there is an open interval containing on which is the largest value, that is, for all x in the

interval.2. A function f is said to have a relative minimum at if

there is an open interval containing on which is the smallest value, that is, for all x in the interval

0x

0( )f x

0( )f x

0x0x

0x

0( ) ( )f x f x

0( ) ( )f x f x

If f has either a relative maximum or a relative minimumat , then f is said to have a relative extremum at 0x 0x

Example

Critical Points

A critical point for a function f to be a point in the domain of f at which either the graph of f has a horizontal tangent line or f is not differentiable.

A critical point is called stationary point of f if ( ) 0f x

Relative extrema and critical pointsSuppose that f is a function defined on an open interval

containing the point 0.x If f has a relative extremum at 0x xthen 0x x is a critical point of f ; that is, either 0( ) 0 or f x f is not differentiable at 0.x

Example Find all critical points of 3( ) 3 1f x x x

Solution 3( ) 3 1f x x x

Differentiating w.r.t x2( ) 3 3f x x

2( ) 3 1f x x

To find critical point setting ( ) 0f x

23 1 0x

2 1x 1x

Example

Find all critical points of

Solution

5 23 3( ) 3 15f x x x

5 23 3( ) 3 15f x x x

Differentiating w.r.t x

( )f x 2

353

3x

132

153x

( )f x 2

35x 1

310x

135x

( 2)x

13

5( 2)x

x

and atAt 2,x ( ) 0f x

0,x ( )f x

Thus 0x and 2x critical points.

does not exist

Locate the critical points and identify which critical points are stationary points.

Example

2

1( )

3

xf x

x

2

1( )

3

d xf x

dx x

2 2

2 2

( 3) ( 1) ( 1) ( 3)

( 3)

d dx x x x

dx dxx

2

22

( 3)(1) ( 1)(2 )

3

x x x

x

Differentiating w.r.t x

2 2

2 2

3 2 2( )

( 3)

x x xf x

x

2

2 2

2 3

( 3)

x x

x

2

2 2

( 2 3)

( 3)

x x

x

2

2 2

3 3)

( 3)

x x x

x

2 2

( 3) 1( 3)

( 3)

x x x

x

2 2

( 3)( 1)

( 3)

x x

x

( ) 0f x

2 2

( 3)( 1)0

( 3)

x x

x

( 3)( 1) 0x x

( 3) 0 orx 1 0x

3 or 1x x

setting

Thus x=1 and x=-3 are stationary points.

3x 1x Stationary points:

First Derivative TestSuppose that f is continuous at a critical point 0.x

1. If ( ) 0f x on an open interval extending left from 0x and

( ) 0f x on an open interval extending right from 0 ,x then

f has a relative maximum at 0.x

2. If on an open interval extending left from ( ) 0f x 0x

0.x0 ,x

and

( ) 0f x on an open interval extending right from thenf has a relative minimum at

3. If ( )f x has same sign on an open interval extending left

from 0x as it does on an open interval extending right from 0 ,x then f does not have a relative extrema 0.x

Use the first derivative test to show that Example

2( ) 3 6 1f x x x has a relative minimum at x=1

2( ) 3 6 1f x x x ( ) 6 6f x x

6( 1)x

f has relative minima at x=1

Solution

x=1 is a critical point as ( ) 0 at 1.f x x

Interval

Test Value

Sing of - +

Conclusionis decreasing on

is decreasing on

1x 1x

c 0 2

( )f c

(0) 6 0f (2) 6 0f ( )f c

f1x

f1x

Second Derivative TestSuppose that f is twice differentiable at the 0.x

(a) If 0( ) 0f x and 0( ) 0,f x then f has relative minimum at 0.x

(b) If 0( ) 0f x and 0( ) 0,f x then f has relative maximum at 0.x

(c) If 0( ) 0f x and 0( ) 0,f x then the test is inconclusive;

that is, f may have a relative maximum, a relative minimum,

or neither at 0.x

ExampleFind the relative extrema of 5 3( ) 3 5f x x x

Solution4 2( ) 15 15f x x x

2 215 ( 1)x x

2( ) 15 ( 1)( 1)f x x x x

3( ) 60 30f x x x 230 (2 1)x x

Critical Points

Setting ( ) 0f x

215 ( 1)( 1) 0x x x 215 0 or 1 0 or 1 0x x x

1,0,1x are critical points

Stationary Point

Second Derivative Test

-30 -has a relative

maximum

0 0 Inconclusive

30 + has a relative

minimum

230 (2 1)x x ( )f x

1x

f

0x

1x

f

Absolute Extrema

Let I be an interval in the domain of a function f. We say that f has an absolute maximum at a point in I if

for all x in I, and we say that f has an absolute minimum at if for all x in I.

0x0x 0( ) ( )f x f x

0( ) ( )f x f x

Extreme value Theorem

If a function f is continuous on a finite closed interval [a, b] then f has both an absolute maximum and an absolute minimum on [a, b].

Procedure for finding the absolute extrema of a continuous function f on a finite closed interval [a, b]

Step 1. Find the critical points of f in (a, b).

Step 2. Evaluate f at all the critical points and at the end points a and b.

Step 3. The largest of the value in step 2 is the absolute maximum value of f on [a, b] and the smallest value is the absolute minimum

Find the absolute maximum and minimum values of the function

Example

3 2( ) 2 15 36f x x x x on the interval [1, 5], and determine where these values occur.

solution2( ) 6 30 36f x x x

2( ) 6( 5 6)f x x x

6( 2)( 3)x x

( ) 0f x at x=2 and x=3

So x=2 and x=3 are stationary points

Evaluating f at the end points, at x=2 and at x=3 and at the endspoints of the interval.

3 2(1) 2(1) 15(1) 36(1) 23f

3 2(2) 2(2) 15(2) 36(2) 28f

3 2(3) 2(3) 15(3) 36(3) 27f

3 2(5) 2(5) 15(5) 36(5) 55f

Absolute minimum is 23 at x=1

Absolute minimum is 55 at x=5