SRqa004

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course Mth163

6/15 6

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

004.

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Question: `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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Your solution:

f(x) = x^2 + 4 when f(3) = 13, f(7) = 53 and f(-5) = 29

I plotted these points and I obtained an exponential graph moving from left to right. The point left is f(-5) = 29 and the point right at f(7) = 53

confidence rating #$&*: 3

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Given Solution:

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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Self-critique (if necessary):

I didn't see the quadratic equation. I'm looking for f(x) = ax^2 + bx + c. I didn't know that b could be 0 and x didn't have to be present to be quadratic.I can see

where the line of symetry is (0,4). If I would have tried the same points like -7 or 5 for f(x) I might have realized it.

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Self-critique rating:

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Question: `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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Your solution:

f(a) = a^2 +4

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

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

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

f(x + h) - f(x) / h

I've tried for an hour or so to solve this last function. I don't feel at all confident in this.

confidence rating #$&*: 1

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Given Solution:

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.

You should have written these expressions out, and the following should probably be represented on your paper in form similar to that given here:

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Self-critique (if necessary):

I missed that f(X) = x^2 +4 and I should substitute this when seeing the f(x) = (x + h) - f(x). On the last problem I wanted to

work out the problem again using ((x + h) - f(x) / h )^2 + 4. I understand it a little better after seeing the answers, and I dentify where I've gone wrong.

I'll practice more.

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Self-critique rating:

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Question: `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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Your solution:

f(x1) = 5x1 + 7

f(x2) = 5x2 + 7

f(x2) - f(x1) / x2 -x1 = 5x2 + 7 - 5x1 + 7 / x2 - x1 = 5 + 7 + 5 + 7 = 24

confidence rating #$&*: 3

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Given Solution:

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.

Compare what you have written down with the expressions below:

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Self-critique (if necessary):

I should have remembered that I have to divide by the entire number (x2 - x1) and I can't decide what to divide by. It makes perfect sense once I see it work out.

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

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Your solution:

f(-2) = -3 = 5(x) + 7

confidence rating #$&*: 3

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Given Solution:

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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