Question

given
$f(t)=int_{0}^{t}\frac{x^{2}+11x + 30}{1+cos^{2}(x)}dx$
at what value of $t$ does the local max of $f(t)$ occur?
$t = $

Explanation:

Step1: Apply Fundamental Theorem of Calculus

$f'(t) = \frac{t^2 + 11t + 30}{1 + \cos^2(t)}$

Step2: Find critical points (set $f'(t)=0$)

The denominator $1+\cos^2(t) \geq 1 > 0$ for all $t$, so solve $t^2 + 11t + 30 = 0$.
Factor: $(t+5)(t+6)=0$, so $t=-5$ or $t=-6$.

Step3: Use sign test for $f'(t)$

• For $t < -6$: $(t+5)(t+6) > 0$, so $f'(t) > 0$
• For $-6 < t < -5$: $(t+5)(t+6) < 0$, so $f'(t) < 0$
• For $t > -5$: $(t+5)(t+6) > 0$, so $f'(t) > 0$

$f'(t)$ changes from positive to negative at $t=-6$, so this is a local maximum.

Answer:

$t=-6$