Thread: 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!

(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.

Originally Posted by HallsofIvy

(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

Originally Posted by jcocker

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

Originally Posted by JaguarXJS

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