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Calculus questions help - Maths SL

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a) Max occurs when f' changes sign from + to - and f'(x)=0, so p=6

b) Look at the graph and points given, f'(2)=-2  (b/c it says f'(x) passes thru (2,-2))

c) Chain rule: [d/dx]g(f(x))=g'(f(x))*f'(x). g(x)=ln(x) => g'(x)=1/x => g'(f(x))=1/f(x). f'(x)=f'(x) so [d/dx]g(f(x))=f'(x)/f(x). They tell you f(2)=3 and f'(2)=-2 from part b, so g'(2)=f(2)/f'(2)=3/-2=-3/2. 

d) I'll start by evaluating the integral give. By the Fundamental Theorem of Calculus I, the integral from a to b of g'(x)dx is = g(b)-g(a). Here, a=2 and b=a, so the integral is = g(a)-g(2). From the problem, we're told that g(x)=ln(f(x)) and that f(2)=3, so g(2)=ln(f(2))=ln(3). So the integral is = g(a)-ln(3). The problem says that ln(3) + the integral = g(a), substitute what we found and you have ln(3)+g(a)-ln(3)=g(a), which simplifies to g(a)=g(a). Verification complete :)

e) Set up an integral: The integral from 2 to 5 of g'(x)dx = B-A   (It would be B+A, but A is negative since it is below the x-axis). The integral evaluates to become g(5)-g(2)=.21-.66. From d) we found that g(2)=ln(3), so we have that g(5)-ln(3)=-.45 => g(5)=ln(3)-.45. You could have also used the formula from d) and substituted a=5, it's the same thing really.

If you have any questions let me know! 

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