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Thread: Past exam question help please...

  1. #1
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    Past exam question help please...

    I am revising for an important exam next week.

    There is one question that I cannot seem to be able to solve (see attached) - I have the answers for it, but not the workings...

    Any help will be much appreciated!
    Attached Thumbnails Attached Thumbnails Past exam question help please...-capture.jpg  
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  2. #2
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    Re: Past exam question help please...

    (a) The crucial point is the 'turning point' (I would say "vertex") of the parabolas. For f it is (2, 3) and for h it is (7, 6). We want to find a and b such that h(x)= f(x+ a)+ b. We want x= 7 in h to correspond to x+ a in f so we want 7+ a= 2. That is, a= -5. f(7- 5)+ b= f(2)+ b= 3+ b= 6 so b= 6- 3= 3.

    (b) Since h(x)= f(x- 5)+ 3, $\displaystyle \int_6^8 h(x)dx= \int_6^8 f(x- 5)+ 3 dx= \int_6^8 f(x- 5) dx+ \int_6^8 3dx$. In the first integral make the change or variable u= x- 5. Then du= dx, f(x- 5)= f(u). When x= 6, u= 6- 5= 1 and when x= 8, u= 8- 5= 3. The integral $\displaystyle \int_6^8 f(x- 5)dx$ becomes $\displaystyle \int_1^3 f(u)du$. Of course, you need to add $\displaystyle \int_6^8 3 dx$.

    (c) h(x)= f(x- 5)+ 3 so, using the chain rule, h'(x)= f'(x- 5) times the derivative of x- 5 which is, of course, 1. h'(x)= f'(x- 5). Unfortunately, since 1 is NOT 8- 5 that doesn't immediately help. However, we can use the symmetry evident in the graph. Because of the symmetry, the slope at x= 8 is the negative of the slope at x= 6 and 6- 5 is 1! h'(6)= f'(6- 5)= 1 so h'(8)= -1.
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  3. #3
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    Re: Past exam question help please...

    Quote Originally Posted by HallsofIvy View Post
    (a) The crucial point is the 'turning point' (I would say "vertex") of the parabolas. For f it is (2, 3) and for h it is (7, 6). We want to find a and b such that h(x)= f(x+ a)+ b. We want x= 7 in h to correspond to x+ a in f so we want 7+ a= 2. That is, a= -5. f(7- 5)+ b= f(2)+ b= 3+ b= 6 so b= 6- 3= 3.

    (b) Since h(x)= f(x- 5)+ 3, $\displaystyle \int_6^8 h(x)dx= \int_6^8 f(x- 5)+ 3 dx= \int_6^8 f(x- 5) dx+ \int_6^8 3dx$. In the first integral make the change or variable u= x- 5. Then du= dx, f(x- 5)= f(u). When x= 6, u= 6- 5= 1 and when x= 8, u= 8- 5= 3. The integral $\displaystyle \int_6^8 f(x- 5)dx$ becomes $\displaystyle \int_1^3 f(u)du$. Of course, you need to add $\displaystyle \int_6^8 3 dx$.

    (c) h(x)= f(x- 5)+ 3 so, using the chain rule, h'(x)= f'(x- 5) times the derivative of x- 5 which is, of course, 1. h'(x)= f'(x- 5). Unfortunately, since 1 is NOT 8- 5 that doesn't immediately help. However, we can use the symmetry evident in the graph. Because of the symmetry, the slope at x= 8 is the negative of the slope at x= 6 and 6- 5 is 1! h'(6)= f'(6- 5)= 1 so h'(8)= -1.
    Thanks a lot!

    Unfortunately the official answer to (c) is -6 though... (b) is 10
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    Re: Past exam question help please...

    Quote Originally Posted by jcocker View Post
    I am revising for an important exam next week.

    There is one question that I cannot seem to be able to solve (see attached) - I have the answers for it, but not the workings...

    Any help will be much appreciated!
    Please do not post on multiple forums
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Past exam question help please...

    Quote Originally Posted by JaguarXJS View Post
    Please do not post on multiple forums
    As I can't find a second copy of this problem on MHF I presume you mean posting on other Help sites. It can be frustrating when people cross sites like this but there are no rules being broken. I don't have a problem with it.

    -Dan
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