course mth163 please coment on my assignment progress xEŞiݒxassignment #002
......!!!!!!!!...................................
20:59:30 `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.
......!!!!!!!!...................................
RESPONSE --> f(3)=3^2+4 f(3)=13 f(7)=7^2+4 f(7)=53 f(-5)=-5^2+4=29 the graph starts in the positive x,y quadrant runs up the positive x ,y quadrant and then drops down to the negative y positive x confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:00:23 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.
......!!!!!!!!...................................
RESPONSE --> same thing self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:06:59 `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.
......!!!!!!!!...................................
RESPONSE --> f(a)=a^2+4 f(x+2)=(x+2)^2+4 f(x+h)=(x+h)^2+4 f(x+h)-f(x)= h^2+4 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:09:36 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.
......!!!!!!!!...................................
RESPONSE --> ok so I should have factored out the expressions self critique assessment: 1
.................................................
......!!!!!!!!...................................
21:19:05 `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.
......!!!!!!!!...................................
RESPONSE --> f(x1)=5(x1)+7 f(x2)=5(x2)+7 (5(x2)+7)-(5x1+7)/(x2-x1) confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:20:52 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.
......!!!!!!!!...................................
RESPONSE --> ok I see where I forgot to factor the - through the second half of the numerator self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:24:31 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
......!!!!!!!!...................................
RESPONSE --> -3=5x+7 -3-7=5x -10=5x -10/5=5x/5 x=-2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:24:58 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
"