Open QA 4

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course Mth 163

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 where f(3)

f = 3^2 +4

f = 13

this is like saying x = 3 and y = 13

(3,13)

f(x) = x^2 + 4 where f(7)

f = 7^2+4

= 53

(7, 53)

f(x) = x^2 + 4 where f(-5)

f = -5^2 + 4

= 29

(-5, 29)

The graph is a parabola with (0,4) as its vertex. Its vertex is facing in the upwards direction so it is also positive.

confidence rating #$&*:

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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 looked at my graph to derive the vertex but this could have been doing using the -b/ (2a) portion of the quadratic formula. We could have found our other two important points by shifting the vertex at (0,4) to the left or right 1.

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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(x)= x^2+4

F(a)= a^2+4

F(x+2) = (x+2)^2+4

= x^2+4x + 8

????How would you have simplified the binomial rather than using the foil method, or is the method the same, we just think of it in a broader context then what traditional foil implies????

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

=x^2+2hx+h^2 +4

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

= x^2 + 2hx + h^2 + 4 -x^2 - 4

= 2hx + h^2

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

= 2x+h

confidence rating #$&*:

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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): didn’t write these out but the representations on the computer are clear to me, sometimes the math becomes enough overwhelming that I need to write it out on paper, but I’ve become comfortable enough with the type writer notation to transport the two formats back and forth, at least up until this point.

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Self-critique rating:

@&

Good.

Most people will need to write this out on paper.

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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) = 5(x1) + 7

= 5x1 + 7

F(x2) = 5(x2) + 7

= 5x2+7

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

= 0

confidence rating #$&*:

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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 thought the x2 and x1 units would cancel out in the division leaving 5-5 = 0, but I see that we remove the common factor of 5, then the units divide out, and we are left with just 5.

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Self-critique rating:

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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:

-3 = 5x+7

-10 = 5x

X = -2

F(-2) = 5(-2) + 7

= -3

confidence rating #$&*:

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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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Self-critique (if necessary): ok

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Self-critique rating: ok

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#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#