Question

exercises 1.3 the fundamental theorem score: 110/260 answered: 11/26 question 12 given \\( f(x) = \int_{0}^{x} \frac{t^2 - 9}{1 + \cos^2(t)} dt \\) at what value of \\( x \\) does the local max of \\( f(x) \\) occur? \\( x = \\) question help: video message instructor submit question jump to answer

Explanation:

Step1: Find derivative of $f(x)$

By the Fundamental Theorem of Calculus Part 1, $f'(x) = \frac{x^2 - 9}{1 + \cos^2(x)}$

Step2: Find critical points

Set $f'(x)=0$. Since $1+\cos^2(x) > 0$ for all $x$, solve $x^2 - 9 = 0$
$x^2 - 9 = 0 \implies x = 3$ or $x = -3$

Step3: Test sign of $f'(x)$

• For $x < -3$: $x^2-9>0$, so $f'(x)>0$ (function increasing)
• For $-3 < x < 3$: $x^2-9<0$, so $f'(x)<0$ (function decreasing)
• For $x > 3$: $x^2-9>0$, so $f'(x)>0$ (function increasing)

A local maximum occurs where $f'(x)$ changes from positive to negative, which is at $x=-3$.

Answer:

$x=-3$