THE DERIVATIVE IN GRAPHING AND APPLICATIONS

ANALYSIS OF FUNCTIONS 1: INCREASING and DECREASING

FUNCTIONS, ROLLE’S THEOREM, MEAN VALUE THEOREM, CONCAVITY

and POINT OF INFLECTION

OBJECTIVES:

• define increasing and decreasing functions;•define concavity and direction of bending that is concave upward or concave downward; •define Rolle’s Theorem and Mean Value Theorem; and • determine the point of inflection.

.

0 2 4

increasing decreasing increasing constant

The term increasing, decreasing, and constant are used to describe the behavior of a function as we travel left to right along its graph. An example is shown below.

INCREASING and DECREASING FUNCTIONS

Definition 4.1.1 (p. 233)

The following definition, which is illustrated in Figure 4.1.2, expresses these intuitive ideas precisely.

Figure 4.1.2 (p. 233)

y

x

Each tangent line has positive slope; function is increasing

y

x

Each tangent line has negative slope;function is decreasing

y

x

Each tangent line has zero slope,function is constant

Theorem 4.1.2 (p. 233)

.gsindecrea and gsinincrea is 3x4xf(x) whichon intervals the Find .1 2 y

2, on increasing is f2x when0x'f,2- on decreasing is f2x when0x'f

thus

2x 24x2x'f

-1

2

3

x

3x4x)x(f 2

increasing

decreasing

EXAMPLE:

.gsindecrea and gsinincrea is xf(x) whichon intervals the Find .2 3

3x)x(f

0, on increasing is f0x when0x'f,0- on increasing is f0x when0x'f

thus

x3x'f 2

increasing

increasing

y

-4

3

4

x-3

ROLLE’S THEOREM AND

THE MEAN-VALUE THEOREM

ROLLE’S THEOREM

This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal.

Theorem 4.8.1 (p. 302)Rolle's Theorem

Figure 4.8.1

EXAMPLE:Find the two x-intercepts of the functionand confirm that f’(c) = 0 at some point between those intercepts.

4x5xxf 2

Solution:

.0cf' which at 1,4

interval the on point a is 25 c so ,

25x ; 05x2xf'

0.cf' that such 1,4 interval the in c point one least atof existence the guaranteed are we Thus .1,4 interval the on satisfied are Theorem sRolle'of hypotheses the ,everywhere

abledifferenti and continuous isf polynomial the since 4x and 1x are intercepts-x the so ,4x1x 4x5x xf 2

1 2 3 4

1

2

-1

-2

x

y

025'f

THE MEAN-VALUE THEOREM Rolle’s Theorem is a special case of a more general result, called the Mean-value Theorem. Geometrically, this theorem states that between any two points A (a,f(a)) and B(b,f(b)) on the graph of a differentiable function f, there is at least one place where the tangent line to the graph is parallel to the secant line joining A and B.(Fig 4.8.5)

Figure 4.8.5 (p. 304)

Note that the slope of the secant line joining A(a,f(a)) and B(b,f(b)) is

abafbfm

and that the slope of the tangent line at c in Figure 4.5.8a is f’(c). Similarly, in Figure 4.5.8b the slopes of the tangent lines at joining A(a,f(a)) and B(b,f(b)) isSince nonvertical parallel lines have the same slope, the Mean-Value Theorem can be stated precisely as follows

.ly respective ,cf' and cf' are c and c 2 12 1

Theorem 4.8.2 (p. 304)Mean-Value Theorem

EXAMPLE:Show that the function satisfies the hypotheses of the mean-value theorem over the interval [0,2], and find all values of c in the interval (0,2) at which the tangent line to the graph of f is parallel to the secant line joining the points (0,f(0)) and (2,f(2)).

1x41xf 3

Solution:

.0,2 interval the in lies 1.15 only ,15.13

3234c Therefore

4c3 0213

4c3

abafbfcf' Thus

4c3cf' and ,

4x3xf' 32fbf ,10faf But

2.b and 0a withd satisfieare Theorem Value- Meanthe of hypotheses the so,0,2 on abledifferenti and 0,2 on continuous is f particular In

. polynomial a is it because everywhere iable different and continuous is f

22

22

1 2 3 4

1

2

-1

-2

x

y 1x41y 3

3

4

CONCAVITY Although the sign of the derivative of f reveals where the graph of f is increasing or decreasing , it does not reveal the direction of the curvature.

Fig. 4.1.8 suggests two ways to characterize the concavity of a differentiable f on an open interval:• f is concave up on an open interval if its tangent lines have increasing slopes on that interval and is concave down if they have decreasing slopes.• f is concave up on an open interval if its graph lies above its tangent lines and concave down if it lies below its tangent lines.

concave up

y

x

concave

down

x

y

increasing slopes decreasing slopes

Figure 4.1.8

Formal definition of the “concave up” and “concave down”.

Definition 4.1.3 (p. 235)

Theorem 4.1.4 (p. 235)

Since the slopes of the tangent lines to the graph of a differentiable function f are the values of its derivative f’,it follows from Theorem 4.1.2 (applied to f’ rather than f )that f’ will be increasing on intervals where f’’ is positive and that f’ will be decreasing on intervals where f’’ is negative. Thus we have the following theorem.

EXAMPLE 1:

0x''f2x''f and 4x2x'f

.,- interval the on up concave is3x4xy function the that suggestsabove figure The 2

y

-1

2

3

x

3x4x)x(f 2

increasing

decreasing

EXAMPLE 2:

0x if 0x''fand 0x if 0x''f

x6x''f and x3x'f

.0, interval the on up concave and .0- interval the on down concave is

xy function the that suggestsabove figure The

2

3

3x)x(f

increasing

increasing

y

-4

3

4

x-3

INFLECTION POINTS Points where the curve changes from concave up to concave down or vice-versa are called points of inflection.

Definition 4.1.5 (p. 236)

Figure 4.1.9 (p. 236)

2-1 3

1

2

-3

y

x

The figure shows the graph of the function . Use the 1st and 2nd derivatives of f to determine the intervals on which f is increasing, decreasing, concave up and concave down. Locate all inflection points and confirm that your conclusions are consistent with the graph.

1x3xxf 23

1x66x6x''f

2xx3x6x3x'f

:SOLUTION2

EXAMPLE 1:

INTERVAL (3x)(x-2) f’(x) CONCLUSIONx<0 (-)(-) + f is increasing on

0<x<2 (+)(-) - f is decreasing on

x>2 (+)(+) + f is increasing on

0,

2,0

,2

INTERVAL (6)(x-1) f’’(x) CONCLUSIONx<1 (-) - f is concave down on

x>1 (+) + f is concave up on ,1

1,

The 2nd table shows that there is a point of inflection at x=1, since f changes from concave up to concave down at that point. The point of inflection is (1,-1).

.xxf of any, if points, inflection the Find 4

up concave ;0x when 0x''f up concave 0;x when0x''f

x12x''f

x4x'f

:SOLUTION

2

3

. 00'f' though even0,x at point inflection no hence andconcavity in change no is there Thus

EXAMPLE 2:

ANALYSIS OF FUNCTIONS 2: RELATIVE EXTREMA; GRAPHING

POLYNOMIALS

OBJECTIVES:

•define maximum, minimum, inflection, stationary and critical points, relative maximum and relative minimum;•determine the critical, maximum and minimum points of any given curve using the first and second derivative tests;•draw the curve using the first and second derivative tests; and•describe the behavior of any given graph in terms of concavity and relative extrema

Definition 4.2.1 (p. 244)

Figure 4.2.1

1x at maximum relative a and 2x and 1x at minima relative a has

1x4xx34x

21y 234

EXAMPLE :

2

-1

3

1

2

-3

y

x-3

3

1

The points x1, x2, x3, x4, and x5 are critical points. Of these, x1, x2, and x5 are stationary points.

Figure 4.2.3 (p. 245)

Figure 4.2.3 illustrates that a relative extremum can also occur at a point where a function is not differentiable. In general, we define 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 (line is vertical). To distinguish between the two types of critical points we call x a stationary point of f if f’(x)=0.

Thus we have the following theorem:

1x3xxf of points critical all Find 3

1 x01x1x01x

,thus

01x1x33x3x'f

:SOLUTION2

EXAMPLE 1:

1x and 1x at occur points critical the Therefore

1x when and -1x when 0xf' that means This

tangent line

tangent line

32

35

x15x3xf of points critical all Find

0 x0x

2x02x ,thus

0x

2x52xx5x10x5x'f

:SOLUTION

31

31

31

31

32

point.ry stationaa is 2x thatand 2x and 0x at occur

points critical the Therefore0x when xf' and 2x when 0xf' that means This

52-1 3

-4

1

-3

y

x

-2

2

1

tangent line

EXAMPLE 2:

FIRST DERIVATIVE TEST

Theorem 4.2.2 asserts that the relative extrema must occur at critical points, but it does not say that a relative extremum occurs at every critical point.

sign.changes f' wherepoints critical those at extremum relative a has function A

Figure 4.2.6 (p. 246)

Theorem 4.2.3 (p. 247) First Derivative Test

•The above theorem simply say that for a continuous function, relative maxima occur at critical points where the derivative changes from (+) to (–) and relative minima where it changes from (–) to (+).

EXAMPLE: 3

235

x15x3xf of points critical the all Find

:SOLUTION

point.y stationara is 2x that and 2x and 0x at occur

points critical The

31

31

31

32

x

2-x5

2xx5

x10x5x'f

that shownhave we

52-1 3

-4

1

-3

y

x

-2

2

1

tangent line

:below shownis derivative this of analysis signA

INTERVAL f’(x)

x<0 (-)/(-) +

0<x<2 (-)/(+) _

x>2 (+)/(+) +

31

x/2x5

imummin relative2x at to from changes f' of signThe imummax relative0x at to from changes f' of signThe

SECOND DERIVATIVE TEST

There is another test for relative extrema that is based on the following geometric observation:

• A function f has a relative maximum at stationary point if the graph of f is concave down on an open interval containing that point.

•A function f has a relative minimum at stationary point if the graph of f is concave up on an open interval containing that point.

Note: The second derivative test is applicable only to stationary points where the 2nd derivative exists.

Figure 4.2.7

EXAMPLE: 35 x5x3xf of extrema relative the Find .1

:SOLUTION

12x30x30x-60xx'' f

1x -1;x 0;x 01x ;01x ;015x

01x1x15x0x' f when

1x1x15x1x15x 15x-15xx' f

23

2

2

22224

STATIONARY POINTS f’’

2nd DERIVATIVE TEST

x=-1 -30 - f has a relative maximum

x=0 0 0 inconclusive

x=1 30 + f has a relative minimum

1x2x30 2

INTERVAL f’(x) Conclusionx<-1 (+)(-)(-) +

x=-1 f has a relative maximum

-1<x<0 (+)(+)(-) -

x=0 f has neither a relative max nor min

0<x<1 (+)(+)(-) -

x=1 f has a relative minimum

x>1 (+)(+)(+) +

1x1xx15 2

2-1

1

y

x

-2

2

1

3xx3y of curve the trace and Analyze .2

1x and -1x 0x1 and 0x1

0x1x13x-13

x33'y

xx3y

2

2

3

0x 0x6

x6''y

:SOLUTION

INTERVAL Conclusion

x<-1 (+)(-)(+)= - + f is decreasing; concave upward

x=-1 -2 0 + f has a relative minimum

-1<x<0 (+)(+)(+)= + + f is increasing; concave upward

x=0 0 3 0 f has a point of inflection

0<x<1 (+)(+)(+)= + - f is increasing; concave downward

x=1 2 0 - f is has a relative maximum

x>1 (+)(+)(-)= - - f is decreasing; concave downward

2x3x

xf

x1x13

x' f

x6 x'' f

-2

2-1

1

y

x

-2

2

1

3x3x3y

2x4x4y of curve the trace and Analyze .3

2x and -2x 0x2 and 0x2

0x4

x2x24x4x44'y

x4x416

x4x8x416'y

x4x2x44x4'y

x4x4y

2222

2

22

2

22

22

22

2

2

32x and 0x

012xx4x8

024x2x44x-

016x4x28x44x-

016x4x42x44x-

016x4x4x4x8x4

x4

x2x4216x4x8x4''y

22

22

222

222

2222

42

222

INTERVAL Conclusion

- - f is decreasing; concave downward

-0.125 0 f has a point of inflection

- + f is decreasing; concave upward

x= -2 -1 0 0.25 f has a relative minimum

-2<x<0 + + f is increasing; concave upward

x=0 0 1 0 f has a point of inflection

0<x<2 + - f is increasing; concave downward

x=2 1 0 -0.25 f is has a relative maximum

- - f is decreasing; concave downward

-0.125 0 f has a point of inflection

- + f is decreasing; concave upward

xf x' f x'' f

32x

32x

2x32

32x2

32x

32x

23

23

-2

2-1

1

y

x

-2

2

1-3-4 3 4

`

2x4x4y

7x12x3x2y .1 23

4x2y .2 2

234 x12x4x3y .3

22

x1xy .4

35 x32x

51y .5

234 x6x8x3y .6

)x2(xy .7 22

2x1xy .8

Exercises:Determine any critical points, point of inflection and trace the curve.