calculus and analytical geometry lecture # 13 mth 104
TRANSCRIPT
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