Given f ( x )

We are given the function:

$f(x) = \int_0^x \frac{t^2 - 25}{1 + \cos^2(t)} \, dt$

To find where the local maximum of $f(x)$ occurs, we need to follow these steps:

Step 1: Compute the derivative of $f(x)$

By the Fundamental Theorem of Calculus, the derivative of an integral function is the integrand evaluated at the upper limit:

$f'(x) = \frac{x^2 - 25}{1 + \cos^2(x)}$

Step 2: Find critical points

The critical points occur when $f'(x) = 0$. So, we set the derivative equal to zero:

$\frac{x^2 - 25}{1 + \cos^2(x)} = 0$

This equation will be zero when the numerator is zero, because the denominator is always positive:

$x^2 - 25 = 0$

$x^2 = 25$

$x = \pm 5$

Step 3: Determine which point is a local maximum

To determine whether $x = 5$ or $x = -5$ corresponds to a local maximum, we examine the second derivative or the sign changes of $f'(x)$. We compute the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx} \left( \frac{x^2 - 25}{1 + \cos^2(x)} \right)$

But instead of calculating $f''(x)$, we can analyze the sign of $f'(x)$ around $x = \pm 5$:

• For $x < -5$, $x^2 - 25$ is positive.
• For $-5 < x < 5$, $x^2 - 25$ is negative.
• For $x > 5$, $x^2 - 25$ is positive.

Therefore, $f'(x)$ changes sign from positive to negative at $x = -5$, indicating a local maximum at $x = -5$.

So, the local maximum of $f(x)$ occurs at:

$x = -5$

Would you like to explore more details on this process or ask any further questions?

Here are 5 questions to deepen understanding:

1. How does the Fundamental Theorem of Calculus apply to integral functions like this?
2. What happens if $x = 5$ is evaluated for a local minimum instead?
3. Why is $f(x)$ continuous and differentiable, and how does this impact finding maxima?
4. Could we use a numerical method to analyze the behavior of the function more closely?
5. How does the second derivative test apply to confirm the nature of critical points?

Tip: Always check sign changes in the derivative to confirm maxima or minima when analyzing critical points.