No Problemmo #4

I'll test each statement, starting with 1: $f'(1)=(1)F'(1)+F(1)$ . Because 1 lies in the interval $(\frac{1}{2},3)$ and $F(1)=0$ , $f'(1)=F'(1)<0$ . Therefore, 1 is correct. Now 2: Since $F(1)=0$ , and $F'(x)<0$ for all $x$ leading up to 2 (and including 2), we can conclude $F(2)<0 \Rightarrow f(2)=2F(2)<0$ . Therefore, 2 is correct. 3 is correct because $x>0$ , $F'(x)<0$ , and $F(x)<0$ (since $F(1)=0$ and the function is decreasing) over the interval $(1,3)$ , which implies $f'(x)=xF'(x)+F(x)<0$ (so, in particular, it can't be equal to zero). 4 is the negation of 3, so it's false. There is not enough information to prove 5. However, there is enough information to prove 7. The only way 5 and 7 can both be true is if $f'(3)=0$ . However, there is not enough information to tell the value of $F'(3)$ , so one cannot tell whether this is true or not. Perhaps someone can contribute to this solution by finding a function that satisfies all of the original hypotheses, but where $f'(3)\neq 0$ . Now, moving to 6: Using $\int_1^3 x^2F'(x)dx=-12$ , and integrating by parts: $\Big[x^2F(x)-\int 2xF(x)dx\Big]_1^3=9F(3)-F(1)-2\int_1^3 xF(x)dx=-12\Rightarrow 9(-4)-0-2\int_1^3 f(x)dx=-12 \Rightarrow \int_1^3 f(x)dx=-12$ . Therefore, 6 is false and 8 is true. The last one to test is 7. Start with $\int_1^3 x^3F''(x)dx=40$ , and integrate by parts: $\Big[x^3F'(x)-\int 3x^2F'(x)dx\Big]_1^3=27F'(3)-F'(1)-3\int_1^3 x^2F'(x)dx$ $=9(3F'(3)+F(3)-F(3))-((1)F'(1)+F(1))-3(-12)=9(f'(3)-F(3))-f'(1)+36$ $=9(f'(3)+4)-f'(1)+36=9f'(3)-f'(1)+72=40\Rightarrow 9f'(3)-f'(1)+32=0$ . Hence, 7 is true. So, the answer is 12378.