Concavity Changes: Analyzing The Second Derivative Of H(x)

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Hey guys! Let's dive into a cool math problem. We're given a function h(x), and we know its second derivative, h''(x) = -(x-1)(3x-9). The problem tells us that h''(x) = 0 when x = 1 and x = 3. Our mission? Figure out where the concavity of h(x) changes. This is super important because it helps us understand the shape of the function – whether it's curving upwards (concave up) or downwards (concave down).

To crack this, we need to understand what the second derivative tells us. The second derivative h''(x) gives us information about the concavity of the original function h(x). Specifically:

  • If h''(x) > 0, the function h(x) is concave up (like a smile).
  • If h''(x) < 0, the function h(x) is concave down (like a frown).
  • If h''(x) = 0, we have a potential point of inflection, where the concavity might change.

So, the key is to analyze the sign of h''(x) in different intervals to see where it flips from positive to negative, or vice versa. This is where the concavity changes happen. Let’s start by analyzing the given second derivative h''(x) = -(x-1)(3x-9). Notice that we already know that h''(x) = 0 when x = 1 and x = 3. These are our critical points, and they divide the number line into three intervals: (-∞, 1), (1, 3), and (3, ∞). We'll test a value within each interval to determine the sign of h''(x).

Understanding the Second Derivative and Concavity

Alright, let's break down this concept of concavity, because it's super important for understanding the shape of a function. The second derivative is like a secret code that tells us whether our function is curving upwards (concave up) or downwards (concave down). Think of it this way: if you're driving a car and the road curves upwards, you're experiencing concave up. If the road curves downwards, you're experiencing concave down. The second derivative h''(x) is the mathematical tool that helps us identify these curves.

Now, how do we actually use the second derivative to determine concavity? Simple! Let's recap:

  • Concave Up: If h''(x) > 0, the function is curving upwards, like a smile (U-shaped).
  • Concave Down: If h''(x) < 0, the function is curving downwards, like a frown (∩-shaped).
  • Point of Inflection: Where h''(x) = 0, we have a potential point of inflection. This is where the concavity might change. It's like a transition point from a smile to a frown, or vice versa.

So, what does this mean for our problem? We know h''(x) = -(x-1)(3x-9). To figure out the intervals where concavity changes, we need to find where h''(x) changes sign. This happens at the points where h''(x) = 0, which we already know are x = 1 and x = 3. These values act as our boundaries, dividing the number line into intervals that we'll test. This is like setting up different test zones to see how h''(x) behaves. In each interval, we are trying to find where concavity changes.

Finding Intervals with Concavity Changes

Let's get down to the nitty-gritty and find those intervals where the concavity of h(x) changes. We've already established our critical points x = 1 and x = 3, and they define our intervals: (-∞, 1), (1, 3), and (3, ∞). Now it's time for some testing. This is the fun part, guys!

Interval (-∞, 1):

Let's pick a test value in this interval. How about x = 0? Now, plug this value into our second derivative: h''(0) = -(0-1)(30-9) = -(-1)(-9) = -9*. Since h''(0) < 0, the function is concave down in this interval. Basically, the original function h(x) is frowning in this interval. That's one interval down.

Interval (1, 3):

Time for another test. Let's try x = 2. Plug that into our second derivative: h''(2) = -(2-1)(32-9) = -(1)(-3) = 3*. Since h''(2) > 0, the function is concave up in this interval. We've got a smiling face in this zone. It's time to test the last interval.

Interval (3, ∞):

Let's use x = 4 as our test value. Substitute this into our second derivative: h''(4) = -(4-1)(34-9) = -(3)(3) = -9*. Because h''(4) < 0, the function is concave down in this interval. Another frown! Alright. To summarize, we have a concave down interval, a concave up interval, and then another concave down interval. The points of inflection are where the concavity changes. In this specific question, the point of inflection changes between the (1,3) interval. So the answer is (1,3).

Identifying the Correct Interval

Okay, guys, we've done the hard work of analyzing the second derivative and determining the concavity in each interval. Now, let's nail down which interval is the answer to the question. Remember, the question asks where the concavity changes. The concavity changes when the sign of the second derivative changes. It's like a traffic light; we are looking for a change in color (sign).

Looking back at our analysis, here's what we found:

  • (-∞, 1): Concave down (h''(x) < 0)
  • (1, 3): Concave up (h''(x) > 0)
  • (3, ∞): Concave down (h''(x) < 0)

See those changes? The concavity changes from concave down to concave up at x = 1, and then changes back to concave down at x = 3. So, the concavity changes within the intervals (1, 3), and then again at (3, ∞). The question asks for one of the intervals where there is a change. Therefore, (1,3) is one of the correct answers. (3, ∞) is another possible answer.

Now, let's look at the multiple-choice options:

(a) (1, 3) (b) (3, ∞) (c) (-∞, -1) (d) (-∞, 3)

Based on our analysis, we know the concavity changes within the intervals (1, 3) and (3, ∞). Therefore, options (a) and (b) are correct. Both of those are the correct answer to this problem! Option (c) is incorrect because the concavity does not change within that interval, and option (d) is incorrect because it includes both concave up and concave down sections, without specifying where the change actually happens. This is a perfect example of how analyzing the second derivative is crucial for understanding the behavior of a function.

Final Thoughts and Key Takeaways

Alright, we've reached the finish line! Let's recap the key takeaways and solidify our understanding of concavity and the second derivative. This is what you should remember:

  1. The Second Derivative's Role: The second derivative, h''(x), tells us about the concavity of the function h(x). It's like a compass guiding us to understand whether the function curves upwards (concave up) or downwards (concave down).
  2. Concavity and Sign:
    • h''(x) > 0 means h(x) is concave up (smiling).
    • h''(x) < 0 means h(x) is concave down (frowning).
  3. Points of Inflection: Points where h''(x) = 0 are potential points of inflection, where the concavity might change.
  4. Interval Analysis: To find intervals where concavity changes, we need to find where h''(x) changes sign. This usually happens at points of inflection, but we need to check the sign of h''(x) on either side of these points.

In our problem, we found that the concavity of h(x) changes within the intervals (1, 3) and (3, ∞). This is because the second derivative, h''(x), changes sign at these points. This knowledge empowers us to sketch the shape of the function h(x), identify its local extrema, and understand its overall behavior. Keep practicing these problems, and you'll become a concavity master in no time! Always remember to break down the problem step-by-step, use test values to understand the sign of the second derivative, and visualize the curve. Keep up the awesome work, guys!