Calculus Made Easy: Finding Critical Points & Function Behavior

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Hey guys! Let's dive into some calculus fun! Today, we're going to explore the function f(x) = 7x² + 4x. Our mission? To find its critical points and figure out where it's increasing or decreasing. Sounds a bit intimidating? Don't sweat it; we'll break it down step by step. We will use the first derivative test, which is a super handy tool in our calculus toolbox.

Understanding the First Derivative Test and the Critical Points

So, what's the deal with the first derivative test? Think of it as a detective for functions. It helps us uncover the function's behavior – where it's going up (increasing) and where it's going down (decreasing). And the most important clues in this investigation are the critical points. These are the special x-values where the function might change its direction. They're like the function's pit stops. The first derivative test works by examining the sign of the first derivative, f'(x), in the intervals around these critical points.

Before we get our hands dirty with calculations, let's clarify what critical points actually are. They are the x-values where either the first derivative, f'(x), is equal to zero, or where f'(x) is undefined. Usually, in polynomial functions like ours, f'(x) is defined everywhere, so we only need to focus on where it equals zero. These points are super important because they often correspond to the local maximums or local minimums of the function (where the function's curve changes direction from increasing to decreasing or vice versa). These points are where the function's slope, or the rate of change, is momentarily zero.

In other words, these are the potential spots where the function might have a peak (maximum) or a valley (minimum). Think of a roller coaster: the critical points are like the top of a hill (maximum) or the bottom of a dip (minimum). The first derivative test helps us determine whether these points are actually local maximums, local minimums, or neither. By analyzing the sign of the derivative on either side of the critical points, we can tell whether the function is increasing or decreasing around these points, which helps us classify the critical points.

Let's get into the actual steps and make this clearer. We'll find the derivative, then find the points where the derivative is zero. Then, we'll analyze the derivative's sign on intervals around these critical points. This will give us a clear picture of the function’s increasing/decreasing behavior.

Finding the Derivative and Critical Points

Alright, let's roll up our sleeves and do some math! First things first, we need to find the derivative of our function f(x) = 7x² + 4x. Remember the power rule? It's our best friend here. Applying the power rule, we get:

f'(x) = 14x + 4

This is the derivative of our function. Now, to find the critical points, we need to find the x-values where f'(x) = 0. So, let's solve the equation:

14x + 4 = 0

Subtracting 4 from both sides gives us:

14x = -4

Dividing both sides by 14, we get:

x = -4/14

Which simplifies to:

x = -2/7

So, we have one critical point at x = -2/7. That's one part of the puzzle solved! This is the potential location where our function could have a local maximum or minimum. We now know one critical point. Let's analyze this with the first derivative test to determine the function's behavior around this point.

Now, the next step is to use the first derivative test to figure out if this critical point is a maximum, a minimum, or neither. It all depends on whether the function is increasing or decreasing on either side of this point. Stay tuned!

Applying the First Derivative Test and Determining Intervals of Increase/Decrease

Okay, we've got our critical point x = -2/7. Now, let's use the first derivative test to see what's happening around this point. The first derivative test involves looking at the sign of f'(x) in the intervals before and after our critical point. This helps us determine if the function is increasing or decreasing in these intervals.

Here’s how we'll do it:

  1. Divide the number line: Our critical point x = -2/7 divides the number line into two intervals: (-∞, -2/7) and (-2/7, ∞).
  2. Choose test values: Pick a test value (any x-value) within each interval. Let’s choose x = -1 for the interval (-∞, -2/7) and x = 0 for the interval (-2/7, ∞).
  3. Evaluate the derivative: Plug these test values into f'(x) = 14x + 4 and check the sign of the result.
    • For x = -1: f'(-1) = 14(-1) + 4 = -10. The derivative is negative.
    • For x = 0: f'(0) = 14(0) + 4 = 4. The derivative is positive.
  4. Interpret the results:
    • Since f'(-1) < 0 in the interval (-∞, -2/7), the function f(x) is decreasing in this interval. It means as we move from left to right, the function's values are going down.
    • Since f'(0) > 0 in the interval (-2/7, ∞), the function f(x) is increasing in this interval. This means as we move from left to right, the function’s values are going up.

So, what does this tell us? The function decreases until x = -2/7 and then starts increasing. This means we have a local minimum at x = -2/7. Imagine a valley. The function goes down to the valley and then heads back up. The critical point is at the bottom of this valley. That is the local minimum.

In summary:

  • The function f(x) is decreasing on the interval (-∞, -2/7).
  • The function f(x) is increasing on the interval (-2/7, ∞).
  • The critical point x = -2/7 is a local minimum.

Conclusion: The Answer

So, let's answer the original question, and choose the right answer! The critical point of the function f(x) = 7x² + 4x is at x = -2/7. And the function is decreasing before this point and increasing after it. The correct answer is therefore:

A) x = -2/7

There you have it! We've successfully navigated the world of derivatives, critical points, and function behavior. We found the critical point using the first derivative, and then used the first derivative test to determine the intervals of increase and decrease. The function decreases until x = -2/7 and then increases, meaning we have a local minimum there. Congrats on making it this far. Calculus might seem tricky at first, but with a bit of patience, it becomes manageable, and even fun!

Keep practicing, and you'll be a calculus whiz in no time! If you want, you can go back and review the steps.