Analyzing F(x) = 3x/(x^2-25): Critical Points & Intervals

Hey guys! Today, we're diving deep into the analysis of the function f(x) = 3x / (x^2 - 25). This might look intimidating at first, but don't worry, we'll break it down step by step. We're going to find its critical values, figure out where it's increasing and decreasing, and pinpoint any local maxima. So, grab your calculators (or your thinking caps!) and let's get started!

(A) Identifying Critical Values of f(x)

First off, what are critical values? Simply put, these are the points where the function's derivative is either zero or undefined. These points are super important because they often mark potential turning points in the graph – where the function switches from increasing to decreasing, or vice versa. They're like the road signs of our function's journey!

To find these critical values, we need to take the derivative of f(x). Remember the quotient rule? It's going to be our best friend here. The quotient rule states that if we have a function f(x) = u(x) / v(x), then its derivative f'(x) is given by:

f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2

In our case, u(x) = 3x and v(x) = x^2 - 25. So, let's find their derivatives:

• u'(x) = 3
• v'(x) = 2x

Now, let's plug these into the quotient rule:

f'(x) = [3(x^2 - 25) - 3x(2x)] / (x^2 - 25)^2

Let's simplify this beast:

f'(x) = [3x^2 - 75 - 6x^2] / (x^2 - 25)^2 f'(x) = [-3x^2 - 75] / (x^2 - 25)^2

Okay, we've got the derivative! Now, we need to find where it's zero or undefined. A fraction is zero when its numerator is zero. So, let's set the numerator equal to zero:

-3x^2 - 75 = 0

Divide both sides by -3:

x^2 + 25 = 0

Subtract 25 from both sides:

x^2 = -25

Now, this is interesting. We're trying to find a real number that, when squared, gives us -25. But, the square of any real number is always non-negative. This means there are no real solutions from this equation. So, the derivative is never zero.

But wait, we're not done yet! We also need to find where the derivative is undefined. A fraction is undefined when its denominator is zero. So, let's set the denominator equal to zero:

(x^2 - 25)^2 = 0

Take the square root of both sides:

x^2 - 25 = 0

Add 25 to both sides:

x^2 = 25

Take the square root of both sides:

x = ±5

Aha! We found our critical values: x = 5 and x = -5. These are the points where the derivative is undefined, which also means the original function f(x) is undefined at these points (we have vertical asymptotes there!). So, our critical values are -5 and 5. Remember these, they're crucial for the next steps!

(B) Determining Intervals Where f(x) is Increasing

Alright, now we need to figure out where our function is increasing. A function is increasing when its derivative is positive. So, we need to find the intervals where f'(x) > 0. Remember our derivative:

f'(x) = [-3x^2 - 75] / (x^2 - 25)^2

Let's analyze the sign of f'(x). The denominator, (x^2 - 25)^2, is always non-negative because it's a square. In fact, it's strictly positive except at x = ±5, where it's zero (but we already know those are critical points).

Now, let's look at the numerator, -3x^2 - 75. This expression is always negative. Why? Because x^2 is always non-negative, so -3x^2 is always non-positive, and subtracting 75 makes it even more negative.

So, we have a negative numerator divided by a positive denominator (except at the critical points). This means that f'(x) is always negative (where it's defined). Therefore, there are no intervals where f(x) is increasing. It's always decreasing or undefined.

(C) Determining Intervals Where f(x) is Decreasing

Great! This one's going to be a breeze after figuring out the increasing intervals. We know that a function is decreasing when its derivative is negative, f'(x) < 0. And guess what? We just established that f'(x) is always negative (except at the critical points where it's undefined).

So, f(x) is decreasing everywhere except at its critical points, x = -5 and x = 5. These are vertical asymptotes, so the function isn't defined there. We need to express the intervals where f(x) is decreasing using interval notation. We'll have three intervals:

• (-∞, -5)
• (-5, 5)
• (5, ∞)

These are the intervals where f(x) is decreasing. We use parentheses because the function is not defined at x = -5 and x = 5.

(D) Listing x-values of Local Maxima

Finally, let's talk about local maxima. A local maximum is a point where the function's value is greater than the values at nearby points. Think of it as a peak in the graph, but not necessarily the highest peak overall.

Local maxima occur where the function changes from increasing to decreasing. But, we already know that f(x) is never increasing! It's always decreasing or undefined. This means that f(x) has no local maxima.

In summary:

• Critical values: -5 and 5
• Intervals where f(x) is increasing: None
• Intervals where f(x) is decreasing: (-∞, -5), (-5, 5), (5, ∞)
• x-values of local maxima: None

We've successfully analyzed the function f(x) = 3x / (x^2 - 25)! We found its critical values, determined where it's increasing and decreasing, and identified its local maxima (or lack thereof). I hope this breakdown was helpful, guys! Keep practicing, and you'll become function analysis masters in no time!