Finding Function Trends: Increasing, Decreasing, Or Constant

Hey math enthusiasts! Let's dive into the awesome world of calculus and figure out how to pinpoint where a function is increasing, decreasing, or chilling out and staying constant. We'll use the power of derivatives to crack the code. Buckle up, because we're about to become function detectives!

Unveiling Function Behavior: The Power of Derivatives

So, what's the deal with derivatives? Think of them as the function's mood ring. They tell us the slope of the function at any given point. If the derivative is positive, the function is going uphill (increasing). If it's negative, the function is going downhill (decreasing). And if it's zero, the function is flat (constant or at a critical point). It's like having a secret weapon to understand how a function behaves!

Before we start, let's brush up on the basics. A function is increasing on an interval if, as you move from left to right along the x-axis, the y-values get bigger. Conversely, a function is decreasing if the y-values get smaller. And if the y-values stay the same, the function is constant. Derivatives help us pinpoint these behaviors by telling us the instantaneous rate of change.

To find these intervals, we'll take the derivative of our function, find the critical points (where the derivative is zero or undefined), and then test the intervals between those critical points to see if the derivative is positive or negative. This is the key to unlocking the secrets of function behavior! Are you guys ready?

First, let's talk about the critical points. Critical points are super important. They're the x-values where the derivative is either equal to zero or doesn't exist. These points are like the function's turning points—where it might change from increasing to decreasing or vice versa. Finding these points is the first step in our investigation.

Next, we'll use a sign chart or test intervals. Once we've found the critical points, we divide the x-axis into intervals using those critical points. Then, we choose a test value within each interval and plug it into the derivative. If the derivative is positive, the function is increasing in that interval. If it's negative, the function is decreasing. If the derivative is zero, the function is constant at that point. This method gives us a clear picture of the function's behavior across different sections of its domain.

So, let's put on our detective hats and get started with some examples! It's going to be fun!

Analyzing h(x) = rac{1}{4}x^4 - 2x^2

Alright, let's analyze the function h(x) = rac{1}{4}x^4 - 2x^2. This is where the magic happens! We'll find where this function is increasing, decreasing, or constant. It's like solving a puzzle, and the derivative is our secret decoder ring!

First things first, we need to find the derivative of $h(x)$, which we'll call $h'(x)$. Using the power rule, we get:

$h'(x) = x^3 - 4x$

Now, we need to find the critical points. These are the x-values where $h'(x) = 0$ or where $h'(x)$ is undefined. Since $h'(x)$ is a polynomial, it's defined everywhere. So, we just need to solve $h'(x) = 0$:

$x^3 - 4x = 0$

Factor out an x:

$x(x^2 - 4) = 0$

Further factor the difference of squares:

$x(x - 2)(x + 2) = 0$

So, the critical points are $x = -2, x = 0$, and $x = 2$. These are the potential turning points of our function, where it might change from increasing to decreasing or vice versa. Remember these points; they're gold!

Next, we'll create a sign chart or use test intervals to determine where $h'(x)$ is positive or negative. We'll choose test values in the intervals created by our critical points: $(-\\infty, -2)$, $(-2, 0)$, $(0, 2)$, and $(2, \\infty)$.

• Interval: $(-\\infty, -2)$: Choose $x = -3$. Then, $h'(-3) = (-3)^3 - 4(-3) = -27 + 12 = -15$. Since $h'(-3) < 0$, $h(x)$ is decreasing on this interval.
• Interval: $(-2, 0)$: Choose $x = -1$. Then, $h'(-1) = (-1)^3 - 4(-1) = -1 + 4 = 3$. Since $h'(-1) > 0$, $h(x)$ is increasing on this interval.
• Interval: $(0, 2)$: Choose $x = 1$. Then, $h'(1) = (1)^3 - 4(1) = 1 - 4 = -3$. Since $h'(1) < 0$, $h(x)$ is decreasing on this interval.
• Interval: $(2, \\infty)$: Choose $x = 3$. Then, $h'(3) = (3)^3 - 4(3) = 27 - 12 = 15$. Since $h'(3) > 0$, $h(x)$ is increasing on this interval.

Therefore, we have:

• $h(x)$ is decreasing on the intervals $(-\\infty, -2)$ and $(0, 2)$.
• $h(x)$ is increasing on the intervals $(-2, 0)$ and $(2, \\infty)$.

Isn't that cool? We've used calculus to pinpoint exactly where this function is going up and down! High five!

Analyzing $g(t) = |t + 2| - |t - 2|$

Now, let's explore $g(t) = |t + 2| - |t - 2|$. This function involves absolute values, which adds a bit of intrigue. We'll use a different method to solve it since the absolute value function is piecewise. We need to break this down into different cases.

First, we need to consider the critical points, which occur where the expressions inside the absolute values change signs. That happens at $t = -2$ and $t = 2$.

• Case 1: $t < -2$: In this case, both $t + 2$ and $t - 2$ are negative. Thus, $|t + 2| = -(t + 2)$ and $|t - 2| = -(t - 2)$. So,

  $g(t) = -(t + 2) - [-(t - 2)] = -t - 2 + t - 2 = -4$. This tells us that $g(t)$ is constant and equal to -4 when $t < -2$.

• Case 2: $-2 \leq t < 2$: In this case, $t + 2$ is non-negative, and $t - 2$ is negative. Thus, $|t + 2| = t + 2$ and $|t - 2| = -(t - 2)$. So,

  $g(t) = (t + 2) - [-(t - 2)] = t + 2 + t - 2 = 2t$. This tells us that $g(t)$ is increasing when $-2 \leq t < 2$.

• Case 3: $t \geq 2$: In this case, both $t + 2$ and $t - 2$ are non-negative. Thus, $|t + 2| = t + 2$ and $|t - 2| = t - 2$. So,

  $g(t) = (t + 2) - (t - 2) = t + 2 - t + 2 = 4$. This tells us that $g(t)$ is constant and equal to 4 when $t \geq 2$.

Now, let's summarize our findings:

• For $t < -2$, $g(t) = -4$, so the function is constant.
• For $-2 < t < 2$, $g(t) = 2t$, which means it's an increasing linear function, so the function is increasing.
• For $t > 2$, $g(t) = 4$, so the function is constant.

Therefore:

• $g(t)$ is constant on the interval $(-\\infty, -2)$ and $(2, \\infty)$.
• $g(t)$ is increasing on the interval $(-2, 2)$.

See? Even with absolute values, we can crack the code! Calculus is like a superpower. The knowledge to use derivatives to analyze functions, is just amazing! Keep exploring, keep questioning, and keep having fun with math, guys!