increasing/decreasing if f ’ (x) > 0 on an interval, then f is increasing. if f ’ (x) > 0...

Increasing/Decreasing Increasing/Decreasing

If f ’ (x) > 0 on an interval, then f If f ’ (x) > 0 on an interval, then f is increasing.is increasing.

If f ’ (x) < 0 on an interval, then f If f ’ (x) < 0 on an interval, then f is decreasing is decreasing

Increasing / Increasing / DecreasingDecreasing

1. Take the derivative of the function, 1. Take the derivative of the function, f’(x)f’(x)

2. Set f’(x) = 02. Set f’(x) = 0

3. Solve for x3. Solve for x


Find all x such that f ’(x) = 0 or Find all x such that f ’(x) = 0 or not continuous {-1.5, not continuous {-1.5, 0.5}0.5}


Find all x such that f ’(x) = 0 or Find all x such that f ’(x) = 0 or not continuous {-1.5, not continuous {-1.5, 0.5}0.5}

Find open intervals on x-axisFind open intervals on x-axis (-1.5, 0.5)(-1.5, 0.5) (-oo, -1.5) or (0.5, +oo) (-oo, -1.5) or (0.5, +oo)


Find open intervals on x-axisFind open intervals on x-axis -oo ---- -1.5 ---- 0.5 ---- +oo-oo ---- -1.5 ---- 0.5 ---- +oo Test f ’(x) in each intervalTest f ’(x) in each interval f’(-2)=3 f’(0)=-2.5 f’(1)=2f’(-2)=3 f’(0)=-2.5 f’(1)=2 + + + 0 - - - - 0 + + ++ + + 0 - - - - 0 + + +


-oo -------- -1.5 --------- 0.5 ------ -oo -------- -1.5 --------- 0.5 ------ +oo+oo

f’(-2)=3 f’(0)=-2.5 f’(1)=2f’(-2)=3 f’(0)=-2.5 f’(1)=2 + + + 0 - - - - - - - 0 + + ++ + + 0 - - - - - - - 0 + + + Increasing Decreasing Increasing Decreasing

IncreasingIncreasing


f’(-2)=3 f’(0)=-2.5 f’(1)=2f’(-2)=3 f’(0)=-2.5 f’(1)=2 Increasing Decreasing Increasing Decreasing

IncreasingIncreasing (-oo,-1.5) (-1.5,0.5) (0.5, +oo)(-oo,-1.5) (-1.5,0.5) (0.5, +oo)

Increasing / Increasing / DecreasingDecreasing1. Take the derivative of the function, f’(x)1. Take the derivative of the function, f’(x)2. Set f’(x) = 02. Set f’(x) = 03. Solve for x3. Solve for xExample – If f(x) = 4/3xExample – If f(x) = 4/3x33+2x+2x22-3x+1-3x+1 f ’(x) = 4xf ’(x) = 4x2 2 + 4x – 3 = 0+ 4x – 3 = 0 (2x - 1)(2x + 3) = 0 so (2x - 1)(2x + 3) = 0 so 2x – 1 = 0 or 2x + 3 = 0 2x – 1 = 0 or 2x + 3 = 0 2x = 1 or 2x = -3 2x = 1 or 2x = -3 x = ½ or x=-1½x = ½ or x=-1½

Increasing / Increasing / DecreasingDecreasing

4. Graph the solutions4. Graph the solutions

5. Test one value of f’(x) in each interval5. Test one value of f’(x) in each interval

6. Test value positive => Increasing6. Test value positive => Increasing

Test value is negative => DecreasingTest value is negative => Decreasing

f(x) = -xf(x) = -x3 3 + 3x+ 3x2 2 + 24x - 32+ 24x - 32

Find all x such that f ’(x) = 0 or Find all x such that f ’(x) = 0 or not continuousnot continuous

y’=-3xy’=-3x22+6x+24 = 0+6x+24 = 0 y’=-3(xy’=-3(x22-2x-8) = 0-2x-8) = 0

f(x) = -xf(x) = -x3 3 + 3x+ 3x2 2 + 24x - 32 + 24x - 32

f ’(x)=-3(xf ’(x)=-3(x2 2 - 2x - 8) = 0- 2x - 8) = 0

-3(x - 4)(x + 2) = 0-3(x - 4)(x + 2) = 0

x – 4 = 0 or x + 2 = 0x – 4 = 0 or x + 2 = 0

x = 4 or x = -2x = 4 or x = -2

f’(x) = -3xf’(x) = -3x2 2 + 6x + 24 + 6x + 24

x = -2 or x = 4x = -2 or x = 4

Set up the intervals test valuesSet up the intervals test values

-oo ------ -2 ------ 4 ------ +oo-oo ------ -2 ------ 4 ------ +oo

f’(-10)<0 f’(0)=24 f’(10)<0f’(-10)<0 f’(0)=24 f’(10)<0

Decreasing Increasing DecreasingDecreasing Increasing Decreasing

(-oo,-2) (-2,4) (4,+oo)(-oo,-2) (-2,4) (4,+oo)

f(x) = xf(x) = x2 2 + 4x + 1+ 4x + 1

f(x) = xf(x) = x2 2 + 4x + 1 find f + 4x + 1 find f ’(x)’(x)

A.A. 2x2x

B.B. xx2 2 + 4+ 4

C.C. 22

D.D. 2x + 42x + 4

Find the x so f ’(x) = 0.Find the x so f ’(x) = 0.When is 2x + 4 = 0 ?When is 2x + 4 = 0 ?

A.A. x = 4x = 4

B.B. x = 2x = 2

C.C. x = -2x = -2

D.D. x = - ½ x = - ½

f ’(x) = 2x + 4 f ’(x) = 2x + 4 -oo - - - - - - - - +oo-oo - - - - - - - - +oo

A.A. f ’(-10) < 0 f ’(10) < 0f ’(-10) < 0 f ’(10) < 0

B.B. f ’(-10) < 0 f ’(10) > 0f ’(-10) < 0 f ’(10) > 0

C.C. f ’(-10) > 0 f ’(10) < 0f ’(-10) > 0 f ’(10) < 0

D.D. f ’(-10) > 0 f ’(10) > 0f ’(-10) > 0 f ’(10) > 0

f ’(-10)=-16 f f ’(-10)=-16 f ’(10)=24’(10)=24A.A. f is increasing on (-oo, +oo)f is increasing on (-oo, +oo)

B.B. f is increasing on (-oo, -2) onlyf is increasing on (-oo, -2) only

C.C. f is increasing on (-2, + oo) onlyf is increasing on (-2, + oo) only

D.D. f is increasing only at x = -2f is increasing only at x = -2

g(x) = x + 1/xg(x) = x + 1/x

g(x) = x + 1/x g’(x) =g(x) = x + 1/x g’(x) =

A.A. -1 – 1/x-1 – 1/x22

B.B. 1 + 1/x1 + 1/x22

C.C. 1 – 1/x1 – 1/x22

g’(x)=1 - 1/xg’(x)=1 - 1/x22 add add fract.fract.= x= x22/x/x22 – 1/x – 1/x22 = =A.A. (x(x2 2 - 1)/ x- 1)/ x22

B.B. (x(x2 +2 + 1)/ x 1)/ x22

C.C. (1 – x(1 – x22)/ x)/ x22

(x(x2 2 - 1)/ x- 1)/ x22 = 0 when = 0 when the numerator does. x the numerator does. x ==A.A. x = 0 or x = 1x = 0 or x = 1

B.B. x = -1 or x = 1x = -1 or x = 1

C.C. x = 0 or x = -1x = 0 or x = -1

g’(x) is not continuous g’(x) is not continuous when x = when x =

2

2

1'( )

xg x

x

g’(x) is not continuous g’(x) is not continuous when x = when x =

0.00.0

0.10.1

2

2

1'( )

xg x

x

Where is f increasing?Where is f increasing?

A.A. (-oo, -1) U (1, +oo)(-oo, -1) U (1, +oo)

B.B. (-oo, -1) U (0, 1)(-oo, -1) U (0, 1)

C.C. (-1, 0) U (0, 1)(-1, 0) U (0, 1)

2

2

1'( )

xg x

x

g has a critical point at g has a critical point at x=c means that g’(c)=0 x=c means that g’(c)=0 or d.n.e.or d.n.e. The critical points The critical points

for g are {-1, 0, for g are {-1, 0, 1}1}

Numerator = 0Numerator = 0 Denominator = 0Denominator = 0

2

2

1'( )

xg x

x

Find the critical points Find the critical points for h(x) = - xfor h(x) = - x3 3 + 12x - 9+ 12x - 9

A.A. x = 3 or -2x = 3 or -2

B.B. x = - 2 or 2x = - 2 or 2

C.C. x = 0 or -3x = 0 or -3

First derivative testFirst derivative test

1. Take the derivative of the function, 1. Take the derivative of the function, f’(x)f’(x)

2. Set f’(x) = 02. Set f’(x) = 0

3. Solve for x – If f(x) = x/(1 + x3. Solve for x – If f(x) = x/(1 + x22))

f ’(x) = f ’(x) =

When x = 1 or x = -1When x = 1 or x = -1

2 2

2 22 2

1 2 10

1 1

x x x x

x x

First derivative testFirst derivative test

1. If f(x) = x + 9/x+2 = x + 9x1. If f(x) = x + 9/x+2 = x + 9x-1-1 + 2 + 2

2. f’(x) = 1 – 9x2. f’(x) = 1 – 9x-2-2 = 1 – 9/x = 1 – 9/x22

3. = 03. = 0

Numerator = 0 when x = 3 or x = -3Numerator = 0 when x = 3 or x = -3

2

2

9x

x

First derivative testFirst derivative test f ’(x) = f ’(x) =

4. 4. 5. 5. f’(-5)= 16/p f’(-1) = -8/p f’(1) = -8/p f’(5)=16/pf’(-5)= 16/p f’(-1) = -8/p f’(1) = -8/p f’(5)=16/p

6. 6. increasing |increasing | decreasing || decreasing | decreasing || decreasing | increasing increasing

relative max @ -3 asymptote min @ 3relative max @ -3 asymptote min @ 3

2

2

9x

x

h(x) = - xh(x) = - x3 3 + 12x - 9+ 12x - 9

h’(x) = -3xh’(x) = -3x22 + 12 = 0 + 12 = 0 12 = 3x12 = 3x22

4 = x4 = x22

-2 = x or 2 = x -2 = x or 2 = x

y = - xy = - x3 3 + 12x - 9 has critical + 12x - 9 has critical points at 2 and -2. Where is points at 2 and -2. Where is y increasing?y increasing?

A.A. (-oo, -2) U (2, +oo)(-oo, -2) U (2, +oo)

B.B. (-2, 2)(-2, 2)

C.C. RR

D.D. (2, -2)(2, -2)

y = - xy = - x3 3 + 12x - 9 has critical + 12x - 9 has critical points at 2 and -2. Where is points at 2 and -2. Where is y decreasing?y decreasing?

A.A. (-oo, -2) U (2, +oo)(-oo, -2) U (2, +oo)

B.B. (-2, 2)(-2, 2)

C.C. RR

D.D. (2, -2)(2, -2)

y = - xy = - x3 3 + 12x – 9 + 12x – 9 Where is the local Where is the local max?max?

y = - xy = - x3 3 + 12x – 9 + 12x – 9 Where is the local Where is the local max?max?

2.02.0

0.10.1

y = - xy = - x3 3 + 12x – 9 + 12x – 9 Where is the local Where is the local min?min?

y = - xy = - x3 3 + 12x – 9 + 12x – 9 Where is the local Where is the local min?min?

-2.0-2.0

0.10.1

Where is f(x) = x/(2xWhere is f(x) = x/(2x22+3)+3)increasing? No increasing? No asymptoteasymptote f ’(x) = [(2xf ’(x) = [(2x2 2 + 3)-x(4x)]/(2x+ 3)-x(4x)]/(2x2 2 + 3)+ 3)22

which is zero when the numerator which is zero when the numerator is.is.

- 2x- 2x2 2 + 3 = 0+ 3 = 0

Where is f(x) = x/(2xWhere is f(x) = x/(2x22+3)+3)increasing? increasing?

- 2x- 2x2 2 + 3 = 0 makes f’(x) = 0+ 3 = 0 makes f’(x) = 0 3 = 2x3 = 2x2 2 or or +-root(3/2) = x+-root(3/2) = x

Where is f(x) = x/(2xWhere is f(x) = x/(2x22+3)+3)increasing? increasing?

- 2x- 2x2 2 + 3 = 0 makes f’(x) = 0+ 3 = 0 makes f’(x) = 0 3 = 2x3 = 2x2 2 or x = +-root(3/2) or x = +-root(3/2)

Where is f(x) = x/(2xWhere is f(x) = x/(2x22+3)+3)increasing? increasing? f ’(x) = (- 2xf ’(x) = (- 2x2 2 + 3)/(2x+ 3)/(2x2 2 + 3)+ 3)22

f ’(-10) = -197/203f ’(-10) = -197/20322<0 <0 f ’(0)=3/9 >0f ’(0)=3/9 >0 f ’(10) = -197/203f ’(10) = -197/20322<0 <0

Where is f(x) = x/(2xWhere is f(x) = x/(2x22+3)+3)increasing? increasing? f ’(x) = (- 2xf ’(x) = (- 2x2 2 + 3)/(2x+ 3)/(2x2 2 + 3)+ 3)22

f ’(-10) = -197/203f ’(-10) = -197/20322<0 <0 f ’(0)=3/9 >0f ’(0)=3/9 >0 f ’(10) = -197/203f ’(10) = -197/20322<0 <0

Answer = ( , ) Answer = ( , ) 3

23

2

Where are the relative Where are the relative max and relative min?max and relative min?

Relative min at x= max at x Relative min at x= max at x = =

3

2 3

2

increasing/decreasing if f ’ (x) > 0 on an interval, then f is increasing. if f ’ (x) > 0...

Documents