course Mth 163 ?|????E??????[~?Q?assignment #004
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09:20:20 `questionNumber 40000 `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(3) = 3^2 + 4 f(3) = 13 f(7) = 7^2 + 4 f(7) = 53 f(-5) = -5^2 + 4 f(-5) = 29 The coordinates for the graph would be (3, 13), (7, 53), and (-5, 29). The graph would be a parabola passing through these points. The vertex would be: -b / 2a = 0 / 2*1 = 0 f(0) = 0 ^2 + 4 = 4. The vertex of the graph would be (0, 4). The points one unit to the left or right of the vertex would be (1, 5) and (1, -5) and would continue on to the points found earlier. confidence assessment: 2
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09:20:37 `questionNumber 40000 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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RESPONSE --> self critique assessment: 3
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09:30:49 `questionNumber 40000 `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 --> We just substitute for each value: f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 = (x+2)(x+2) + 4 = x(x+2)+2(x+2) + 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+h)^2 + 4 - f(x) = x^2 + 2xh + h^2 + 4 - f(x) [f(x+h) - f(x)] / h = [x^2 + 2xh + h^2 + 4 - f(x)] / h = x^2 / h + 2x + h + 4/h - f(x)/h confidence assessment: 2
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09:35:29 06-19-2008 09:35:29 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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NOTES -------> For the last two expression I did not put the expression for f(x) in to solve. Therefore, it looks like I just did a couple steps and stopped before solving the entire function. Need to remember when it has an expression like f(x+h) - f(x) that I have to substitute for both f(x) values into the expression.
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09:35:30 `questionNumber 40000 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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RESPONSE --> self critique assessment: 2
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09:45:13 `questionNumber 40000 `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 --> Now we just substitute each expression into f(x) = 5x + 7: f(x1) = 5x1 + 7 f(x2) = 5x2 + 7 [f(x2) - f(x1)] / (x2 - x1) = [ 5x2 + 7 - ( 5x1 + 7)] / (x2 - x1) = ( 5x2 - 5x1 ) / (x2 - x1) = We can factor out 5 and get 5 ( x2 - x1) / (x2 - x1) = 5 confidence assessment: 2
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09:46:00 06-19-2008 09:46:00 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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NOTES -------> I still need to practice the expressions like the third that has all the parts to it.
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09:46:01 `questionNumber 40000 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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RESPONSE --> self critique assessment: 2
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09:49:02 `questionNumber 40000 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> If we want to find what value of x is f(x) = -3 we would write the expression like this: 5x + 7 = -3 We would just solve for this function to get the value: 5x + 7 - 7 = -3 - 7 5x = -10 x = -2 confidence assessment: 2
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09:49:31 `questionNumber 40000 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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RESPONSE --> self critique assessment: 3
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