Analyzing G(x): Intervals, Extrema, And Inflection Points

Hey math enthusiasts! Let's dive into a cool calculus problem. We're going to break down the function g(x) = 150 + 8x^3 + x^4. Our mission? To uncover its secrets, specifically focusing on its increasing and decreasing intervals, local extrema (maximums and minimums), and inflection points. Buckle up, because we're about to explore the fascinating world of this polynomial function, step by step! This breakdown is designed to be super clear, so even if calculus isn't your everyday thing, you'll be able to follow along. We will go through the process to understand the behavior of the given function. Let's get started!

(a) Unveiling Intervals of Increase and Decrease

Alright, first things first: let's figure out where our function g(x) is going up (increasing) and where it's going down (decreasing). To do this, we need to bring in the concept of the derivative. The derivative, g'(x), tells us the slope of the function at any given point. If g'(x) is positive, the function is increasing; if it's negative, the function is decreasing. The derivative of g(x) = 150 + 8x^3 + x^4 is g'(x) = 24x^2 + 4x^3.

Now, a critical point is where the derivative equals zero or is undefined. In our case, g'(x) is a polynomial, so it's defined everywhere. Thus, we need to find where g'(x) = 0. This is where the magic happens; solving 24x^2 + 4x^3 = 0 gives us the critical points. Let's factor out 4x^2: 4x^2(6 + x) = 0. This means x = 0 or x = -6 are our critical points. These points are super important because they're potential spots where the function changes direction.

Next, we'll create a sign chart (or interval test) to see what's happening around these critical points. We divide the number line into intervals based on our critical points: (-∞, -6), (-6, 0), and (0, ∞). Then, we pick a test value within each interval and plug it into g'(x) to see if the derivative is positive or negative. For (-∞, -6), let's use x = -7: g'(-7) = 24(-7)^2 + 4(-7)^3 = 1176 - 1372 = -196. This is negative, meaning g(x) is decreasing in this interval. For (-6, 0), let's use x = -1: g'(-1) = 24(-1)^2 + 4(-1)^3 = 24 - 4 = 20. This is positive, so g(x) is increasing here. For (0, ∞), let's use x = 1: g'(1) = 24(1)^2 + 4(1)^3 = 24 + 4 = 28. This is positive, so g(x) is increasing here.

So, to sum it up: g(x) is decreasing on the interval (-∞, -6) and increasing on the intervals (-6, 0) and (0, ∞). We've successfully navigated the first part of our mission! Isn't calculus fun? This means the function decreases until it reaches x = -6, and then it starts increasing. When it reaches x = 0, the increase continues. Now, let's move on to explore the other aspects of the function.

(b) Pinpointing Local Minimums and Maximums

Now, let's move on to the local extrema. These are the peaks (local maximums) and valleys (local minimums) of our function. We already have the critical points from part (a), which are potential locations for these extrema. We know that the function changes direction at these points. To classify them, we can use the first derivative test (which we have already used) or the second derivative test. Let's use the first derivative test. We already found out that g(x) decreases on (-∞, -6) and increases on (-6, 0). This means that at x = -6, the function changes from decreasing to increasing, creating a local minimum. Specifically, the local minimum occurs at x = -6, but what is the value of this minimum?

To find the value, we plug x = -6 into the original function g(x): g(-6) = 150 + 8(-6)^3 + (-6)^4 = 150 - 1728 + 1296 = -282. So, the local minimum value is -282. Now, let's analyze what's happening around x = 0. The function increases on both sides of x = 0, on (-6, 0) and (0, ∞). This means that x = 0 is neither a local minimum nor a local maximum. It's a point where the function changes its rate of increase. The function flattens out here, but it doesn't change direction. Think of it like a brief pause before continuing the ascent. Therefore, g(x) has a local minimum at x = -6 with a value of -282 and no local maximums. That means that the function's value decreases until it reaches -6, where the value is -282, then it increases indefinitely. This is the magic of understanding derivatives; the information they carry provides an understanding of a function's behavior.

(c) Uncovering Inflection Points

Lastly, let's hunt for those inflection points. These are the points where the concavity of the function changes – where it switches from curving upwards (concave up) to curving downwards (concave down), or vice versa. To find inflection points, we need to look at the second derivative, g''(x). The second derivative tells us about the concavity. If g''(x) > 0, the function is concave up; if g''(x) < 0, it's concave down. First, we need to find g''(x). We know that g'(x) = 24x^2 + 4x^3. So, taking the derivative again, we get g''(x) = 48x + 12x^2.

Now, we need to find where g''(x) = 0 or is undefined. Again, g''(x) is a polynomial and defined everywhere. So, let's solve 48x + 12x^2 = 0. Factoring out 12x, we get 12x(4 + x) = 0. This means x = 0 or x = -4. These are our potential inflection points. We need to check the sign of g''(x) in the intervals determined by these points: (-∞, -4), (-4, 0), and (0, ∞). For (-∞, -4), let's use x = -5: g''(-5) = 48(-5) + 12(-5)^2 = -240 + 300 = 60. This is positive, meaning the function is concave up. For (-4, 0), let's use x = -1: g''(-1) = 48(-1) + 12(-1)^2 = -48 + 12 = -36. This is negative, meaning the function is concave down. For (0, ∞), let's use x = 1: g''(1) = 48(1) + 12(1)^2 = 48 + 12 = 60. This is positive, meaning the function is concave up.

So, we see that the concavity changes at x = -4 and x = 0. Therefore, x = -4 and x = 0 are inflection points. To find the y-values of these inflection points, plug the x-values into the original function. For x = -4: g(-4) = 150 + 8(-4)^3 + (-4)^4 = 150 - 512 + 256 = -106. For x = 0: g(0) = 150 + 8(0)^3 + (0)^4 = 150. Therefore, the inflection points are (-4, -106) and (0, 150). We have now successfully found everything we need to understand the function. Congrats on following along! You are now equipped with the knowledge to describe the function's characteristics. Remember that understanding the function's derivative is key to interpreting its behavior. Keep practicing, and you'll become a calculus pro in no time! Calculus is a tool for understanding the world, from the behavior of a curve to the movement of a planet. It’s all connected, and it's all fascinating!