# Thread: Local Maxima of a definite integral...

1. ## Local Maxima of a definite integral...

I have such a question: - at what value of t does the local maxima of f(t) occur?

$\displaystyle f(t) = \displaystyle \int_{0}^{t} \left( \frac{x^2 +10x + 24}{1 + cos^2(x)} \right)dx$

so, my attempts in the base of the part 1 of the FTC I have

$\displaystyle f'(t) = \left( \frac{t^2 +10t + 24}{1 + cos^2(t)} \right)dx$ , since the denominator would never be $\displaystyle = 0$ I evaluated the numerator

which gave me two critical points $\displaystyle x_1 = -4; x_2 = -6$, and evaluating the f''(t) at those points gave me (-4) > 0 , and local maxima, but it actually is considered wrong in my homework,

what did I miss here?

2. ## Re: Local Maxima of a definite integral...

Never mind. I was looking as derivative not overall function.

-Dan

It seems that I can't remove the attached image. Ah well.

3. ## Re: Local Maxima of a definite integral...

Originally Posted by dokrbb
I have such a question: - at what value of t does the local maxima of f(t) occur?

$\displaystyle f(t) = \displaystyle \int_{0}^{t} \left( \frac{x^2 +10x + 24}{1 + cos^2(x)} \right)dx$

so, my attempts in the base of the part 1 of the FTC I have

$\displaystyle f'(t) = \left( \frac{t^2 +10t + 24}{1 + cos^2(t)} \right)dx$ , since the denominator would never be $\displaystyle = 0$ I evaluated the numerator

which gave me two critical points $\displaystyle x_1 = -4; x_2 = -6$, and evaluating the f''(t) at those points gave me (-4) > 0 , and local maxima, but it actually is considered wrong in my homework,

what did I miss here?
Maxima are indicated by $\displaystyle f''(t) < 0$, OP.

4. ## Re: Local Maxima of a definite integral...

This is not my day!

-Dan

5. ## Re: Local Maxima of a definite integral...

I got it guys, it is actually the another one $\displaystyle x = -6$ at wich it gives $\displaystyle f''(x) < 0$... why I was looking for the >0 values...,