For Which Interval(s) Is The Function Increasing And Decreasing?$y=\left|2x^2 + X - 6\right|$A. Increasing For \[$-2 \ \textless \ X \ \textless \ 0\$\] And \[$x \ \textgreater \ \frac{3}{2}\$\]; Decreasing For \[$x \

Analyzing the Increasing and Decreasing Intervals of a Function

In mathematics, understanding the behavior of a function is crucial for various applications, including optimization, modeling real-world phenomena, and solving equations. One essential aspect of function analysis is identifying the intervals where the function is increasing or decreasing. In this article, we will delve into the increasing and decreasing intervals of the function $y=\left|2x^2 + x - 6\right|$.

The given function is an absolute value function, which means it can be expressed as the absolute value of a quadratic expression. The absolute value function is defined as:

$$y = \left|f(x)\right| = \begin{cases} f(x) & \text{if } f(x) \geq 0 \\ -f(x) & \text{if } f(x) < 0 \end{cases}$$

In our case, the quadratic expression is $2x^2 + x - 6$. To analyze the increasing and decreasing intervals, we need to find the critical points of the function, which occur when the derivative of the function is equal to zero or undefined.

To find the critical points, we first need to find the derivative of the function. Using the chain rule, we get:

$$\frac{dy}{dx} = \frac{d}{dx} \left|2x^2 + x - 6\right| = \frac{d}{dx} \left(2x^2 + x - 6\right) \cdot \text{sign}(2x^2 + x - 6)$$

where $\text{sign}(x)$ is the sign function, which returns $1$ if $x \geq 0$ and $-1$ if $x < 0$.

Now, we need to find the derivative of the quadratic expression $2x^2 + x - 6$. Using the power rule, we get:

$$\frac{d}{dx} \left(2x^2 + x - 6\right) = 4x + 1$$

So, the derivative of the function is:

$$\frac{dy}{dx} = (4x + 1) \cdot \text{sign}(2x^2 + x - 6)$$

To find the critical points, we need to set the derivative equal to zero or undefined. Since the derivative is a product of two functions, we need to consider both cases.

Case 1: Derivative is zero

Setting the derivative equal to zero, we get:

$$(4x + 1) \cdot \text{sign}(2x^2 + x - 6) = 0$$

Since the sign function is either $1$ or $-1$, we can rewrite the equation as:

$$4x + 1 = 0 \quad \text{or} \quad 4x + 1 = 0$$

Solving for $x$, we get:

$$x = -\frac{1}{4} \quad \text{or} \quad x = -\frac{1}{4}$$

However, this is a repeated root, which means that the derivative is zero at $x = -\frac{1}{4}$.

Case 2: Derivative is undefined

The derivative is undefined when the denominator of the sign function is zero, which occurs when $2x^2 + x - 6 = 0$. Solving for $x$, we get:

$$x = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}$$

So, the critical points are:

$$x = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2$$

Now that we have found the critical points, we can analyze the increasing and decreasing intervals of the function.

To determine the increasing and decreasing intervals, we need to examine the sign of the derivative in each interval. We can do this by choosing a test point in each interval and evaluating the derivative at that point.

Interval 1: $x < -2$

Choosing a test point $x = -3$, we get:

$$\frac{dy}{dx} = (4(-3) + 1) \cdot \text{sign}(2(-3)^2 + (-3) - 6) = (-11) \cdot \text{sign}(15 - 3 - 6) = (-11) \cdot \text{sign}(6) = (-11) \cdot 1 = -11$$

Since the derivative is negative, the function is decreasing in this interval.

Interval 2: $-2 < x < -\frac{1}{4}$

Choosing a test point $x = -1$, we get:

$$\frac{dy}{dx} = (4(-1) + 1) \cdot \text{sign}(2(-1)^2 + (-1) - 6) = (-3) \cdot \text{sign}(2 - 1 - 6) = (-3) \cdot \text{sign}(-5) = (-3) \cdot (-1) = 3$$

Since the derivative is positive, the function is increasing in this interval.

Interval 3: $-\frac{1}{4} < x < \frac{3}{2}$

Choosing a test point $x = 0$, we get:

$$\frac{dy}{dx} = (4(0) + 1) \cdot \text{sign}(2(0)^2 + 0 - 6) = (1) \cdot \text{sign}(-6) = (1) \cdot (-1) = -1$$

Since the derivative is negative, the function is decreasing in this interval.

Interval 4: $x > \frac{3}{2}$

Choosing a test point $x = 2$, we get:

$$\frac{dy}{dx} = (4(2) + 1) \cdot \text{sign}(2(2)^2 + 2 - 6) = (9) \cdot \text{sign}(8 + 2 - 6) = (9) \cdot \text{sign}(4) = (9) \cdot 1 = 9$$

Since the derivative is positive, the function is increasing in this interval.

In conclusion, the function $y=\left|2x^2 + x - 6\right|$ is increasing in the intervals $-2 < x < -\frac{1}{4}$ and $x > \frac{3}{2}$, and decreasing in the intervals $x < -2$ and $-\frac{1}{4} < x < \frac{3}{2}$.
Q&A: Increasing and Decreasing Intervals of a Function

In our previous article, we analyzed the increasing and decreasing intervals of the function $y=\left|2x^2 + x - 6\right|$. In this article, we will answer some frequently asked questions related to the increasing and decreasing intervals of a function.

Q: What is the difference between an increasing and decreasing function?

A: An increasing function is a function that gets larger as the input variable increases. On the other hand, a decreasing function is a function that gets smaller as the input variable increases.

Q: How do I determine the increasing and decreasing intervals of a function?

A: To determine the increasing and decreasing intervals of a function, you need to find the critical points of the function, which occur when the derivative of the function is equal to zero or undefined. Then, you need to examine the sign of the derivative in each interval to determine whether the function is increasing or decreasing.

Q: What is a critical point?

A: A critical point is a point on the graph of a function where the derivative of the function is equal to zero or undefined. Critical points can be used to determine the increasing and decreasing intervals of a function.

Q: How do I find the critical points of a function?

A: To find the critical points of a function, you need to take the derivative of the function and set it equal to zero or undefined. Then, you need to solve for the input variable to find the critical points.

Q: What is the significance of the sign of the derivative?

A: The sign of the derivative determines whether the function is increasing or decreasing. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing.

Q: Can a function have multiple increasing and decreasing intervals?

A: Yes, a function can have multiple increasing and decreasing intervals. For example, the function $y=\left|2x^2 + x - 6\right|$ has four increasing and decreasing intervals.

Q: How do I use the increasing and decreasing intervals to solve problems?

A: The increasing and decreasing intervals of a function can be used to solve a variety of problems, including optimization problems, modeling real-world phenomena, and solving equations.

Q: What are some common mistakes to avoid when analyzing the increasing and decreasing intervals of a function?

A: Some common mistakes to avoid when analyzing the increasing and decreasing intervals of a function include:

• Not finding all the critical points of the function
• Not examining the sign of the derivative in each interval
• Not considering the endpoints of the intervals
• Not using the increasing and decreasing intervals to solve problems

In conclusion, the increasing and decreasing intervals of a function are an essential concept in calculus. By understanding how to determine the increasing and decreasing intervals of a function, you can solve a variety of problems and gain a deeper understanding of the behavior of functions.

• Q: What is the difference between an increasing and decreasing function?
• A: An increasing function is a function that gets larger as the input variable increases. On the other hand, a decreasing function is a function that gets smaller as the input variable increases.
• Q: How do I determine the increasing and decreasing intervals of a function?
• A: To determine the increasing and decreasing intervals of a function, you need to find the critical points of the function, which occur when the derivative of the function is equal to zero or undefined. Then, you need to examine the sign of the derivative in each interval to determine whether the function is increasing or decreasing.
• Q: What is a critical point?
• A: A critical point is a point on the graph of a function where the derivative of the function is equal to zero or undefined. Critical points can be used to determine the increasing and decreasing intervals of a function.
• Q: How do I find the critical points of a function?
• A: To find the critical points of a function, you need to take the derivative of the function and set it equal to zero or undefined. Then, you need to solve for the input variable to find the critical points.
• Q: What is the significance of the sign of the derivative?
• A: The sign of the derivative determines whether the function is increasing or decreasing. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing.
• Q: Can a function have multiple increasing and decreasing intervals?
• A: Yes, a function can have multiple increasing and decreasing intervals. For example, the function $y=\left|2x^2 + x - 6\right|$ has four increasing and decreasing intervals.
• Q: How do I use the increasing and decreasing intervals to solve problems?
• A: The increasing and decreasing intervals of a function can be used to solve a variety of problems, including optimization problems, modeling real-world phenomena, and solving equations.
• Q: What are some common mistakes to avoid when analyzing the increasing and decreasing intervals of a function?
• A: Some common mistakes to avoid when analyzing the increasing and decreasing intervals of a function include:
  • Not finding all the critical points of the function
  • Not examining the sign of the derivative in each interval
  • Not considering the endpoints of the intervals
  • Not using the increasing and decreasing intervals to solve problems