Lynxarin Posted May 9, 2016 Report Share Posted May 9, 2016 Hi, can anyone explain how to solve the following Maths SL calculus questions? Reply Link to post Share on other sites More sharing options...
TheNintendoChip Posted May 9, 2016 Report Share Posted May 9, 2016 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! 1 Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.