2012 pre-test. 1. Find the area of the region bounded by y=x+3, the x-axis, x=3 and x=5. | | | 1 2 3...

2012 pre-test

1. Find the area of the region bounded by y=x+3, the x-axis, x=3 and x=5.

| | | | | | 1 2 3 4 5

y = x

5 -

4 -

3 -

2 -

1 -

0

y = x + 3

2

3

6

5

8

(6 + 8)

2

2

2. Find the equation of the line with slope 2 that intersects the x-axis where x=3.

5 -

4 -

3 -

2 -

1 -

0| | | | | | 1 2 3 4 5

y = m x + b

2

y = 2* x + b

0 3

- 6

3. The parabola y = x2 has slope 6 at the point (3, 9). Thus, the tangent line to the curve at this point has slope 6. Find all points where the line perpendicular to this tangent line intersects the parabola. 50 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y = m* x + b

6

y = * x + b-16

3

99

9.5

y = * x + 9.5-16

y = * x + 9.5

-16

y = x2 x2

= * x + 9.5

-16

x = 3 (given)

x = -3.167

1( )

6y f x

x

(1), ( 5)f f

(2 ) (2) (2 ), and

f h ff h

h

4. Let . Find ,

(simplify)

y = 1x + 6

1

y = 1x + 6

-5

y = 1x + 6

2+h

y = 18 + h 1

8 + h - 1

2 + 6

h

=

18 + h

- 12 + 6

h =

= (8 + h) 8 88(8 + h)

-

h

- h- 1

1

- 18(8 + h)

2( ) 1y f x x (0), (2)f f

(3 ) (3) (3 ), and

f h ff h

h

5. Let . Find

(simplify)

f( x ) = x2 + 1 00

2 f( 0 ) = 1

f( x ) = x2 + 1 22

2 f( 2 ) = 5

f( x ) = x2 + 1

f = x2 + 1

(3 + h)(3 + h)

2

2 f( 3 + h ) = h2 + 6h +10

f (3 + h) - f (3)

h

2( ) 1y f x x (0), (2)f f

(3 ) (3) (3 ), and

f h ff h

h

5. Let . Find

(simplify)

2 f( 0 ) = 1

2 f( 2 ) = 5 f = x2 + 1

2

2 f( 3 + h ) = h2 + 6h +10

= f (3 + h) - f (3)

h

h2 + 6h +10

f(3) = 32 + 1

10

-

h

6. Consider . Rationalize the numerator.

9 3h

h

9 3h

h

=

9 + h - 9

a2 - b2 = (a-b)(a+b)

a - b a + b

h ( )

+1

(3), (1), ( 1) and ( 3)f f f f

23 2( )

2

xf x

x

7. Evaluate

if

.

f 3( ) 5

f 1( ) 1

f 1( ) 0.333

f 3( ) 1

2( ) 1 on 3

( ) 2 on 3

f x x x

g x ax x

If

and f and g meet at x=3, find a.

f(3) = 32 -1

g(3) = 2a*3

= 8

= 6a

8

= 6a

a = 8/6

9. Find the inverse of: 2

( )3 5

xf x

x

2( )

3 5

xf x

x

y 3x + 5(3x + 5)(3x + 5) y = x + 2

3xy + 5y = x + 25y -x-

x(3y - 1) = 2 - 5y

3y -1 3y -1

2 - 5y

3 y -1x = y

x

x

2 - 5x

3 x -1y =

10. Find the remainder when is divided by x2 + 2

( ): (x2 + 2) = x

x*(x2 + 2) = x3 + 2x x3 + 2xsubtract

- 6

- 6*(x2 + 2) = - 6x2 - 12 - 6x2 - 12

subtract

( ) r = +8

11. Sketch the set of graphs by building up from the first and using scaling and translations: (label vertex & intercepts)

2 2

2 2

22

; 4 ( 3) ;

1( 3) ; 4 ( 3) ;

41

( 3) ; 4 2( 3) ;4

y x y x

y x y x

y x y x


y = x2 12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =(x - 3)2

horizontalshift

plot y = (x - 3)2


y = (x-3) 2

12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

expand/shrink

y = - (x - 3)2

plot y = - (x - 3)2


y = x2 12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y = 4 - (x - 3)2

horizontalshift

verticalshift

shrink/expand

plot y = 4 - (x - 3)2


y = 4 - (x - 3)2

12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 ( 4 - (x - 3)2)

shrink/expand

plot y = ¼ (4 - (x - 3)2)


y = x2 12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 (4 - [2(x - 3)]2

horizontalshift

plot y = ¼ (4 - (2x - 3)2)


y = (x - 3)2 12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 (4 - [2(x - 3)]2

shrink/expand

plot y = ¼ (4 - (2x - 3)2)


y = [2(x - 3)]2 12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 (4 - [2(x - 3)]2

reverse

plot y = ¼ (4 - (2x - 3)2)


y =-[2(x - 3)]2

12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 (4 - [2(x - 3)]2

vertical shift

plot y = ¼ (4 - (2x - 3)2)


y =4 - [2(x - 3)]2

12.5 -

10 -

7.5 -

5 -

2.5 -

0| | | | | | 1 2 3 4 5

y =1/4 (4 - [2(x - 3)]2

expand/shrink

plot y = ¼ (4 - (2x - 3)2)

12. A circular piece of paper with a 4 inch radius has a sector with arc length x cut from it. (See below). The edges of the remaining figure are joined to make a cone. Find the circumference of the base of the cone, the radius of the cone and the height of the cone all in terms of x.

x

R

12. A circular piece of paper with a 4 inch radius has a sector with arc length x cut from it. (See below). The edges of the remaining figure are joined to make a cone.a. Find the circumference of the base of the cone.

R

2r = 2R-x

ranswer is 2R-x x

R

x

R

12. A circular piece of paper with a 4 inch radius has a sector with arc length x cut from it. (See below). The edges of the remaining figure are joined to make a cone.b. Find the radius of the cone

x

RR

r

2r = 2R-x

answer is r = 2R - x

2

12. A circular piece of paper with a 4 inch radius has a sector with arc length x cut from it. (See below). The edges of the remaining figure are joined to make a cone.a. Find the height of the cone

x

RR

r

h

r = 2R - x

2r2 + h2 = R2( )2

answer is h= R2 - 2R-x

2( )2

2R - x

2

13. Find the roots of

and all values of x where the function lies above the x-axis.

3 2

2

4 3( )

( 2)( 4)

x x xf x

x x

x3 - 4x2 + 3x

(x-2)(x+4)2

x(x2 - 4x1 + 3)

(x-2)(x+4)2

x(x - 1)(x - 3)

(x-2)(x+4)2

( )2 always positiveat x = 2 and x = -4f(x) doesn’t exist

| | | | | |0 1 2 3 4 5

x

x -3x -2x -1

= =

red = positiveblue = negativepositive = above x axis

x<0 1<x<2 3<x

roots 0,1,3

-4

x≠-4 x≠2

14. Find all values of x satisfying both algebraically and graphically and explain what the region means in terms of distance.

3 4x

5 -

4 -

3 -

2 -

1 -

0 | | | | | | | | 1 2 3 4 5 6 7

y =

x

- 3

y = - x

+3

answer - 1 < x < 7

14. Find all values of x satisfying both algebraically and graphically and explain what the region means in terms of distance.

3 4x

if x - 3 > 0

x - 3 < 4

- (x - 3) < 4

if x - 3 < 0

x < 7 x > - 1

3+ +3 - x + 3 < 4 - 3- 3 - x < 1

- 1 < x < 7

15. Sketch y = - sin(2x) on -2 ≤ x ≤ 2

-22

1

-1

sin(x)

shrinking

sin(2x)

reverse

- sin(2x)

16. Solve for x: 2ln(x - 3) - ln(x +15) = ln3 2ln(x - 3) - ln(x +15) = ln3

= ln(a) - ln(b)a

bln ( )

ln(an) = n*ln(a)

ln[(x-3)2] - ln(x+15) = ln(3)

(x - 3)2

x + 15 = 3(x +15) (x +15)3x + 45- 3x - 45 + - 3x - 45 x2 - 6x + 9 - 3x - 45 = 0

x2 - 9x - 36 = 0 x1= -3, x2 = 12

(x - 3)2

x + 15ln ( ) = ln(3)

domain of the function x > 3

17. Find the area of the square

7

3

a

a2 =

72

3

2-area = = 40 area = 40

18. An open top box is constructed from a rectangular piece of cardboard with dimensions 14 in x 22 in by cutting equal squares with side x from the corners and folding the sides up. Express the volume of the resulting box in terms of x.

2214

- 2x

- 2x

x

V = (22 - 2x)(14 - 2x)x

(x - 3)2

x + 15ln ( ) = ln(3)

2012 pre-test. 1. Find the area of the region bounded by y=x+3, the x-axis, x=3 and x=5. | | | 1 2 3...

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Transcript of 2012 pre-test. 1. Find the area of the region bounded by y=x+3, the x-axis, x=3 and x=5. | | | 1 2 3...