Find Local Max of f(x) w/ FTC2

[irok]

Homework Statement
Question One:
Find a continuous function f and a number a such that

2 + \int_{a}^{x} \frac {f(t)} {t^{6}} \,dt = 6 x^{-1}

Question Two:
At what value of x does the local max of f(x) occur?
f(x) = \int_0^x \frac{ t^2 - 25 }{ 1+\cos^2(t)} dt

The attempt at a solution
I just need some pointers of where to get started.
Question One:

So I used FTC1 on both sides,

2 + f(x) / x^{6} = 6x^{-1}

f(x)= 6x^{5} - 2

I'm not sure how to find a, evaluation theorem?

Question Two:

 

[badphysicist]

I don't know about an analytic solution, but the second part of the problem is very feasible numerically. You can solve in Mathematica in only a few lines by turning it into a minimization problem.

 

[irok]

Well, for Question one:

Can anyone confirm that f(x) = 6x^{5} and a = 2.

I'm pretty sure that a = 2 since,

F(x) - F(a) = [ 6x^{5} / x^{6} ] - [ 2 ] = 6x^{-1} - 2

 

[matt grime]

How does the FTC just let you drop an integral sign out like that? (In 1.)

 

[HallsofIvy]

Homework Statement
Question One:
Find a continuous function f and a number a such that

2 + \int_{a}^{x} \frac {f(t)} {t^{6}} \,dt = 6 x^{-1}

Question Two:
At what value of x does the local max of f(x) occur?
f(x) = \int_0^x \frac{ t^2 - 25 }{ 1+\cos^2(t)} dt

The attempt at a solution
I just need some pointers of where to get started.
Question One:

So I used FTC1 on both sides,

2 + f(x) / x^{6} = 6x^{-1}

No, that is not correct. You have differentiated the left side of the equation but not the right.

f(x)= 6x^{5} - 2

I'm not sure how to find a, evaluation theorem?

Question Two:

Once you have found the function, put it into the integeral.