asst_4

course Mth 163

Assignment 4 exercises1. Of course you know, or at least suspect, that your function models won't always be quadratic. Obtain the indicated expressions for the following functions, none of which are quadratic:

Where f(x) = x^3, find f(-2), f(2), f(-a), f(a), f(x-4) and f(x) - 4. f(-2) = (-2)^3 = -8 f(2) = (2)^3 = 8 f(-a) = (-a)^3 = -a^3 f(a) = (a)^3 = a^3 f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4) = [x(x-4)-4(x-4)](x-4) = [x^2 - 4x - 4x + 16] (x-4) = [x^2 -8x + 16] (x-4) = x(x^2 - 8x + 16) -4(x^2 - 8x + 16) = x^3 - 8x^2 + 16x - 4x^2 + 32x - 64 = x^3 - 12x^2 + 48x - 64 f(x) - 4 = (x)^3 - 4 = x^3 - 4 Where f(x) = 2^x, find f(-2), f(2), f(-a), f(a), f(x+3) and f(x) + 3 f(-2) = 2^(-2) = 1/4 f(2) = 2^(2) = 4 f(-a) = 2^(-a) = 1/(2^a) f(a) = 2^(a) = 2^a f(x+3) = 2^(x+3) f(x)+3 = 2^(x) + 3 = 2^x + 3 2. Obtain expressions for the following: Where value(t) = $1000 (1.07)^t value(0) = $1000 (1.07)^0 = $1000 (1) = $1000 value(1) = $1000 (1.07)^1 = $1000(1.07) = $1070 value(2) = $1000 (1.07)^2 = $1000(1.1449) = $1144.90 value(t+3) = $1000 (1.07)^(t+3) value(t+3)/value(t) = $1000 (1.07)^(t+3)/$1000(1.07)^t, subtract exponents when dividing and the base is the same = $1000 (1.07)^(t+3-t), finish subtracting the exponent = $1000 (1.07)^3, simplify exponent = $1000(1.225043), multiply = $1225.04 Where illumination(distance) = 50 / distance^2: illumination(1) = 50 / 1^2 = 50/1 = 50 illumination(2) = 50 / 2^2 = 50/4 = 25 illumination(3) = 50 / 3^2 = 50/9 = 5.55... illumination(distance)/illumination(2*distance) = (50/distance^2)/(50/distance^2), multiply by the reciprocal = (50/distance^2)*[(2distance)^2/50], multiply = [50(2distance)^2]/[50(distance)^2], 50's cancel out = (2distance)^2/distance^2, distribute exponent = 4distance^2/distance^2

note that your final expression reduces to just 4; distance^2 / distance^2 is 1.

3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Use your graph to estimate the following: The value of x for which f(x) = 60. x = 3 The value f(7). y = 30 The difference between f(7) and f(9). 30-27 = 3 The difference in x values between the points where f(x) = 70 and where f(x) = 30. 25 - 9 = -6.5 4. If a temperature vs. clock time function is given by y = temperature = T(t), then what is the symbolic expression for each of the following: The temperature at time t = 3. y = temp = T(3) The temperature at time t = 5. y = temp = T(5) The change in temperature between t = 3 and t = 5. T(3) - T(5) The average of the temperatures at t = 3 and t = 5. [T(3)+T(5)]/2 What equation would we solve to find the clock time when the model predicts a temperature of 150? 150=T(t) How would we go about finding the length of time required for the temperature to fall from 80 to 30? 80=T(t) 30=T(t) 80 = a 30 = b a-b = time for temp. to fall. 5. Questions about your depth vs. time model For your model of depth vs. time, based on in-class measurements, answer the following questions and give detailed reasons for your answers as well as sketches. Use the f(x) notation at every opportunity as you give and reason out your answers: For how long was the depth between 34 and 47 centimeters? 70s - 51s = 19sec. By how much did the depth change between t = 23 seconds and t = 34 seconds? 70cm-60cm=10cm On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds? (70-60cm)/(23-34s) = - 10cm / 11s = approx. -1cm/s On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? about -1 cm/s 6. A hypothetical depth vs. time model based on three points, none of which are actual data points Sketch a graph of the following depth vs. time data: ( 0 sec, 96 cm) (10 sec, 89 cm) (20 sec, 68 cm) (30 sec, 65 cm) (40 sec, 48 cm) (50 sec, 49 cm) (60 sec, 36 cm) (70 sec, 41 cm) These data were obviously taken by someone with either bad instruments or a high degree of incompetence. However, on the average they might well give a good quadratic model. Sketch the data and sketch a smooth curve that doesn't touch any data points but comes as close as possible to the data points, on the average. Your curve will go pretty much through the middle of the data set. Pick three points on this curve, approximately equally spaced in the y direction, and use them as a basis for constructing a function model. (10,50) (30,60) (60,40) y = ax^2 + bx + c 80 = a(10)^2 + b(10) + c => 100a + 10b + c = 80 (`1) 60 = a(30)^2 + b(30) + c => 900a + 30b + c = 60 (`2) 40 = a(60)^2 + b(60) + c =>3600a + 60b + c = 40 (`3) `1 100a + 10b + c = 80 -`2 900a + 30b + c = 60 ------------------------ `4 -800a - 20b = 20 `1 100a + 10b + c = 80 -`3 3600a + 60b + c = 40 ------------------------- `5 -35000a - 50b = 40 50* `4 -40000a - 1000b = 1000 20*-`5 -70000a - 1000b = 800 --------------------------- -30000a = 200 a = -0.00666... `4 -800(0.00666...)-20b=20 -5.333...-20b=20 -20b=25.3333333... b=-1.2666666... `1 100(0.00666...) + 10(-1.26666...) + c = 80 0.666...-12.666666... + c = 80 -12 + c = 80 c = 92 y = 0.006666...x^2 -1.266666...x + 92 sec cm prediction(cm) deviation 0 96 92 4 10 89 80 9 20 68 69.333... -1.333... 30 65 60 5 40 48 82 -4 50 49 45.333... 3.666... 60 36 40 -4 70 41 36 5 Determine the average deviation for your model, and graph your function. average deviation 17.3333... How close is your model to the curve you sketched earlier? very close How well does your function seem to model the data? fits well Week 2 Quiz#1 Version 11 Find vertex and zeros of y = 0.02t^2 +(-1.8)t + 76 a= .02 b= -1.8 c=76 xVertex= 1.8/(2*.02) = 1.8/0.04 = 45 yVertex= .02(45)^2-1.8(45)+76 = .02(2025)-1.8(45)+76 = 40.5-81+76 = 35.5 Vertex: (45, 35.5) x = (1.8+- sqrt(1.8^2-4*.02*76))/(2*.02) = (1.8+- sqrt(3.24 - 6.08))/0.04 There are no zeros, the graph does not cross the x axis. ÃBêá›æçÊžíÕ¡‹ÕDým¬—¼eÀ±• Student Name: assignment #004

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21:11:01 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.

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RESPONSE --> f(x) = x^2 + 4 f(3) = 3^2 + 4 = 9+4 = 13 f(7) = 7^2 + 4 = 49 + 4 = 53 f(-5) = (-5)^2 + 4 = 25+4 = 29 (3,13) (7,53) (-5,29) The graph looks like it could possibly be a parabola with a vertex of (0,4) when a = 1 b = 0 c = 4 The points 1 unit to the left and right would be (1,5) and (-1,5)

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21:11:04 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.

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21:23:12 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.

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RESPONSE --> f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 = (x+2)(x+2)+4 = x^2 + 4x +4 +4 =x^2+4x+8 f(x+h) = (x+h)^2 + 4 =(x+h)(x+h)+4=x^2+2xh+h^2+4 f(x+h) - f(x)= (x^2+2xh+h^2+4) - (x^2+4) = 2xh+h^2 [f(x+h) - f(x)]/h = (2xh+h^2)/h = 2x+h

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21:23:35 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.

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21:33:21 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.

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RESPONSE --> f(x1) = 5(x1) + 7 f(x2) = 5(x2) + 7 [ f(x2) - f(x1) ] / ( x2 - x1 ) = [ (5*x2+7) - (5*x1+7) ] / ( x2 - x1 ), subtract sevens = [ (5*x2) - (5*x1) ] / ( x2 - x1 ), do reverse distributive property to take out five = [ 5(x2-x1)]/(x2-x1) = 5

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21:33:28 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.

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21:35:56 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?

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RESPONSE --> f(x) = 5x + 7, substiture -3 in for f(x) -3 = 5x +7, subtract 7 from both sides -10 = 5x, divide both sides by 5 -2 = x

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21:35:58 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.

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"

Your work is excellent. Let me know if you have questions.

Note that it isn't necessary for you to give a complete solution to every assigned problem. The Query program is designed to save you the considerable time required to key in the complete solutions; it will ask you questions about the work you have already done on paper. You're welcome to submit complete solutions if you wish, especially on questions not covered by the Query program, but that is optional.

asst_4

course Mth 163

Assignment 4 exercises1. Of course you know, or at least suspect, that your function models won't always be quadratic. Obtain the indicated expressions for the following functions, none of which are quadratic:

Where f(x) = x^3, find f(-2), f(2), f(-a), f(a), f(x-4) and f(x) - 4.

f(-2) = (-2)^3 = -8

f(2) = (2)^3 = 8

f(-a) = (-a)^3 = -a^3

f(a) = (a)^3 = a^3

f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4)

= [x(x-4)-4(x-4)](x-4)

= [x^2 - 4x - 4x + 16] (x-4)

= [x^2 -8x + 16] (x-4)

= x(x^2 - 8x + 16) -4(x^2 - 8x + 16)

= x^3 - 8x^2 + 16x - 4x^2 + 32x - 64

= x^3 - 12x^2 + 48x - 64

f(x) - 4 = (x)^3 - 4 = x^3 - 4

Where f(x) = 2^x, find f(-2), f(2), f(-a), f(a), f(x+3) and f(x) + 3

f(-2) = 2^(-2) = 1/4

f(2) = 2^(2) = 4

f(-a) = 2^(-a) = 1/(2^a)

f(a) = 2^(a) = 2^a

f(x+3) = 2^(x+3)

f(x)+3 = 2^(x) + 3 = 2^x + 3

2. Obtain expressions for the following:

Where value(t) = $1000 (1.07)^t

value(0) = $1000 (1.07)^0 = $1000 (1) = $1000

value(1) = $1000 (1.07)^1 = $1000(1.07) = $1070

value(2) = $1000 (1.07)^2 = $1000(1.1449) = $1144.90

value(t+3) = $1000 (1.07)^(t+3)

value(t+3)/value(t) = $1000 (1.07)^(t+3)/$1000(1.07)^t, subtract exponents when dividing and the base is the same

= $1000 (1.07)^(t+3-t), finish subtracting the exponent

= $1000 (1.07)^3, simplify exponent

= $1000(1.225043), multiply

= $1225.04

Where illumination(distance) = 50 / distance^2:

illumination(1) = 50 / 1^2 = 50/1 = 50

illumination(2) = 50 / 2^2 = 50/4 = 25

illumination(3) = 50 / 3^2 = 50/9 = 5.55...

illumination(distance)/illumination(2*distance) = (50/distance^2)/(50/distance^2), multiply by the reciprocal

= (50/distance^2)*[(2distance)^2/50], multiply

= [50(2distance)^2]/[50(distance)^2], 50's cancel out

= (2distance)^2/distance^2, distribute exponent

= 4distance^2/distance^2

note that your final expression reduces to just 4; distance^2 / distance^2 is 1.

3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25.

Use your graph to estimate the following:

The value of x for which f(x) = 60. x = 3

The value f(7). y = 30

The difference between f(7) and f(9). 30-27 = 3

The difference in x values between the points where f(x) = 70 and where f(x) = 30. 25 - 9 = -6.5

4. If a temperature vs. clock time function is given by y = temperature = T(t), then what is the symbolic expression for each of the following:

The temperature at time t = 3. y = temp = T(3)

The temperature at time t = 5. y = temp = T(5)

The change in temperature between t = 3 and t = 5. T(3) - T(5)

The average of the temperatures at t = 3 and t = 5. [T(3)+T(5)]/2

What equation would we solve to find the clock time when the model predicts a temperature of 150? 150=T(t)

How would we go about finding the length of time required for the temperature to fall from 80 to 30?

80=T(t) 30=T(t)

80 = a 30 = b

a-b = time for temp. to fall.

5. Questions about your depth vs. time model

For your model of depth vs. time, based on in-class measurements, answer the following questions and give detailed reasons for your answers as well as sketches. Use the f(x) notation at every opportunity as you give and reason out your answers:

For how long was the depth between 34 and 47 centimeters? 70s - 51s = 19sec.

By how much did the depth change between t = 23 seconds and t = 34 seconds? 70cm-60cm=10cm

On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?

(70-60cm)/(23-34s) = - 10cm / 11s = approx. -1cm/s

On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? about -1 cm/s

6. A hypothetical depth vs. time model based on three points, none of which are actual data points

Sketch a graph of the following depth vs. time data:

( 0 sec, 96 cm)

(10 sec, 89 cm)

(20 sec, 68 cm)

(30 sec, 65 cm)

(40 sec, 48 cm)

(50 sec, 49 cm)

(60 sec, 36 cm)

(70 sec, 41 cm)

These data were obviously taken by someone with either bad instruments or a high degree of incompetence. However, on the average they might well give a good quadratic model.

Sketch the data and sketch a smooth curve that doesn't touch any data points but comes as close as possible to the data points, on the average. Your curve will go pretty much through the middle of the data set.

Pick three points on this curve, approximately equally spaced in the y direction, and use them as a basis for constructing a function model.

(10,50) (30,60) (60,40)

y = ax^2 + bx + c

80 = a(10)^2 + b(10) + c => 100a + 10b + c = 80 (`1)

60 = a(30)^2 + b(30) + c => 900a + 30b + c = 60 (`2)

40 = a(60)^2 + b(60) + c =>3600a + 60b + c = 40 (`3)

`1 100a + 10b + c = 80

-`2 900a + 30b + c = 60

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`4 -800a - 20b = 20

`1 100a + 10b + c = 80

-`3 3600a + 60b + c = 40

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`5 -35000a - 50b = 40

50* `4 -40000a - 1000b = 1000

20*-`5 -70000a - 1000b = 800

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-30000a = 200

a = -0.00666...

`4 -800(0.00666...)-20b=20

-5.333...-20b=20

-20b=25.3333333...

b=-1.2666666...

`1 100(0.00666...) + 10(-1.26666...) + c = 80

0.666...-12.666666... + c = 80

-12 + c = 80

c = 92

y = 0.006666...x^2 -1.266666...x + 92

sec cm prediction(cm) deviation

0 96 92 4

10 89 80 9

20 68 69.333... -1.333...

30 65 60 5

40 48 82 -4

50 49 45.333... 3.666...

60 36 40 -4

70 41 36 5

Determine the average deviation for your model, and graph your function. average deviation 17.3333...

How close is your model to the curve you sketched earlier? very close

How well does your function seem to model the data? fits well

Week 2 Quiz#1

Version 11

Find vertex and zeros of y = 0.02t^2 +(-1.8)t + 76

a= .02 b= -1.8 c=76

xVertex= 1.8/(2*.02)

= 1.8/0.04

= 45

yVertex= .02(45)^2-1.8(45)+76

= .02(2025)-1.8(45)+76

= 40.5-81+76

= 35.5

Vertex: (45, 35.5)

x = (1.8+- sqrt(1.8^2-4*.02*76))/(2*.02)

= (1.8+- sqrt(3.24 - 6.08))/0.04

There are no zeros, the graph does not cross the x axis.

ÃBêá›æçÊžíÕ¡‹ÕDým¬—¼eÀ±•

Student Name:

assignment #004

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21:11:01

`q001. Note that this assignment has 4 questions

If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.

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RESPONSE -->

f(x) = x^2 + 4

f(3) = 3^2 + 4 = 9+4 = 13

f(7) = 7^2 + 4 = 49 + 4 = 53

f(-5) = (-5)^2 + 4 = 25+4 = 29

(3,13) (7,53) (-5,29)

The graph looks like it could possibly be a parabola with a vertex of (0,4) when a = 1 b = 0 c = 4

The points 1 unit to the left and right would be (1,5) and (-1,5)

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21:11:04

f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain

f(3) = 3^2 + 4 = 9 + 4 = 13.

Similarly we have

f(7) = 7^2 + 4 = 49 + 4 = 53 and

f(-5) = (-5)^2 + 9 = 25 + 4 = 29.

Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4.

The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.

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21:23:12

`q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.

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RESPONSE -->

f(a) = a^2 + 4

f(x+2) = (x+2)^2 + 4 = (x+2)(x+2)+4 = x^2 + 4x +4 +4 =x^2+4x+8

f(x+h) = (x+h)^2 + 4 =(x+h)(x+h)+4=x^2+2xh+h^2+4

f(x+h) - f(x)= (x^2+2xh+h^2+4) - (x^2+4) = 2xh+h^2

[f(x+h) - f(x)]/h = (2xh+h^2)/h = 2x+h

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21:23:35

If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a:

f(a) = a^2 + 4.

Similarly to find f(x+2) we replace x with x + 2:

f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8.

To find f(x+h) we replace x with x + h to obtain

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4.

To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h):

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that

[ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.

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21:33:21

`q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.

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RESPONSE -->

f(x1) = 5(x1) + 7

f(x2) = 5(x2) + 7

[ f(x2) - f(x1) ] / ( x2 - x1 )

= [ (5*x2+7) - (5*x1+7) ] / ( x2 - x1 ), subtract sevens

= [ (5*x2) - (5*x1) ] / ( x2 - x1 ), do reverse distributive property to take out five

= [ 5(x2-x1)]/(x2-x1)

= 5

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21:33:28

Replacing x by the specified quantities we obtain the following:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

We can factor 5 out of the numerator to obtain

5 ( x2 - x1 ) / ( x2 - x1 ) = 5.

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21:35:56

`q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?

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f(x) = 5x + 7, substiture -3 in for f(x)

-3 = 5x +7, subtract 7 from both sides

-10 = 5x, divide both sides by 5

-2 = x

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21:35:58

If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation

5x + 7 = -3.

We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.

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