Consider The Function F ( X ) = 2 X 3 + 6 X 2 − 144 X + 11 F(x) = 2x^3 + 6x^2 - 144x + 11 F ( X ) = 2 X 3 + 6 X 2 − 144 X + 11 , Where { -6 \ \textless \ X \ \textless \ 5$}$.This Function Has An Absolute { \square$}$ Value Equal To { \square$}$ At { X = \square$}$.

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Analyzing the Given Function and Its Absolute Value

Introduction

In this article, we will delve into the analysis of a given function, f(x)=2x3+6x2−144x+11f(x) = 2x^3 + 6x^2 - 144x + 11, and its absolute value within the specified interval, −6≤x≤5-6 \leq x \leq 5. The absolute value of a function is a measure of its distance from zero on the number line, and it plays a crucial role in various mathematical applications, including optimization problems and data analysis.

Understanding the Function

The given function, f(x)=2x3+6x2−144x+11f(x) = 2x^3 + 6x^2 - 144x + 11, is a cubic polynomial, which means it has a degree of three. This implies that the function will have a graph that is a cubic curve, with a possible turning point or inflection point. To analyze the function, we need to find its critical points, which are the values of xx where the function's derivative is equal to zero or undefined.

Finding the Derivative

To find the derivative of the function, we will apply the power rule of differentiation, which states that if f(x)=xnf(x) = x^n, then f′(x)=nxn−1f'(x) = nx^{n-1}. Using this rule, we can find the derivative of the given function:

f′(x)=ddx(2x3+6x2−144x+11)f'(x) = \frac{d}{dx}(2x^3 + 6x^2 - 144x + 11)

f′(x)=6x2+12x−144f'(x) = 6x^2 + 12x - 144

Finding the Critical Points

Now that we have the derivative of the function, we can find the critical points by setting the derivative equal to zero and solving for xx:

6x2+12x−144=06x^2 + 12x - 144 = 0

We can simplify this equation by dividing both sides by 6:

x2+2x−24=0x^2 + 2x - 24 = 0

This is a quadratic equation, and we can solve it using the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a=1a = 1, b=2b = 2, and c=−24c = -24. Plugging these values into the formula, we get:

x=−2±22−4(1)(−24)2(1)x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)}

x=−2±4+962x = \frac{-2 \pm \sqrt{4 + 96}}{2}

x=−2±1002x = \frac{-2 \pm \sqrt{100}}{2}

x=−2±102x = \frac{-2 \pm 10}{2}

This gives us two possible values for xx:

x=−2+102=4x = \frac{-2 + 10}{2} = 4

x=−2−102=−6x = \frac{-2 - 10}{2} = -6

Finding the Absolute Value

Now that we have the critical points, we can find the absolute value of the function at these points. The absolute value of a function is defined as:

∣f(x)∣={f(x)if f(x)≥0−f(x)if f(x)<0|f(x)| = \begin{cases} f(x) & \text{if } f(x) \geq 0 \\ -f(x) & \text{if } f(x) < 0 \end{cases}

We can evaluate the function at the critical points to determine its sign:

f(−6)=2(−6)3+6(−6)2−144(−6)+11f(-6) = 2(-6)^3 + 6(-6)^2 - 144(-6) + 11

f(−6)=−432+216+864+11f(-6) = -432 + 216 + 864 + 11

f(−6)=659f(-6) = 659

Since f(−6)>0f(-6) > 0, the absolute value of the function at x=−6x = -6 is simply f(−6)f(-6):

∣f(−6)∣=f(−6)=659|f(-6)| = f(-6) = 659

Similarly, we can evaluate the function at x=4x = 4:

f(4)=2(4)3+6(4)2−144(4)+11f(4) = 2(4)^3 + 6(4)^2 - 144(4) + 11

f(4)=128+96−576+11f(4) = 128 + 96 - 576 + 11

f(4)=−341f(4) = -341

Since f(4)<0f(4) < 0, the absolute value of the function at x=4x = 4 is the negative of f(4)f(4):

∣f(4)∣=−f(4)=341|f(4)| = -f(4) = 341

Conclusion

In conclusion, the absolute value of the function f(x)=2x3+6x2−144x+11f(x) = 2x^3 + 6x^2 - 144x + 11 is equal to 659 at x=−6x = -6 and 341 at x=4x = 4. These values are the maximum and minimum values of the function within the specified interval, −6≤x≤5-6 \leq x \leq 5. The absolute value of a function plays a crucial role in various mathematical applications, including optimization problems and data analysis.
Q&A: Analyzing the Given Function and Its Absolute Value

Introduction

In our previous article, we analyzed the given function, f(x)=2x3+6x2−144x+11f(x) = 2x^3 + 6x^2 - 144x + 11, and its absolute value within the specified interval, −6≤x≤5-6 \leq x \leq 5. In this article, we will answer some frequently asked questions related to the analysis of the function and its absolute value.

Q1: What is the significance of the absolute value of a function?

A1: The absolute value of a function is a measure of its distance from zero on the number line. It plays a crucial role in various mathematical applications, including optimization problems and data analysis.

Q2: How do you find the absolute value of a function?

A2: To find the absolute value of a function, you need to evaluate the function at a given point and determine its sign. If the function is positive, the absolute value is the function itself. If the function is negative, the absolute value is the negative of the function.

Q3: What are the critical points of the function?

A3: The critical points of the function are the values of xx where the function's derivative is equal to zero or undefined. In this case, the critical points are x=−6x = -6 and x=4x = 4.

Q4: What is the absolute value of the function at the critical points?

A4: The absolute value of the function at x=−6x = -6 is 659, and the absolute value of the function at x=4x = 4 is 341.

Q5: How do you determine the maximum and minimum values of a function within a specified interval?

A5: To determine the maximum and minimum values of a function within a specified interval, you need to evaluate the function at the critical points and the endpoints of the interval. In this case, the maximum value of the function is 659 at x=−6x = -6, and the minimum value of the function is 341 at x=4x = 4.

Q6: What is the significance of the absolute value in optimization problems?

A6: The absolute value plays a crucial role in optimization problems, as it allows us to find the maximum and minimum values of a function within a specified interval.

Q7: How do you use the absolute value in data analysis?

A7: The absolute value is used in data analysis to find the distance between data points and the mean or median of the data. It is also used to identify outliers in the data.

Q8: Can you provide an example of how to use the absolute value in a real-world problem?

A8: Yes, consider a company that wants to minimize its costs. The company's costs can be represented by a function, and the absolute value of the function can be used to find the minimum cost. In this case, the absolute value would represent the distance between the company's costs and the minimum cost.

Conclusion

In conclusion, the absolute value of a function plays a crucial role in various mathematical applications, including optimization problems and data analysis. By understanding the absolute value of a function, we can determine its maximum and minimum values within a specified interval and use it to solve real-world problems.

Frequently Asked Questions

  • Q: What is the absolute value of a function? A: The absolute value of a function is a measure of its distance from zero on the number line.
  • Q: How do you find the absolute value of a function? A: To find the absolute value of a function, you need to evaluate the function at a given point and determine its sign.
  • Q: What are the critical points of the function? A: The critical points of the function are the values of xx where the function's derivative is equal to zero or undefined.
  • Q: What is the absolute value of the function at the critical points? A: The absolute value of the function at x=−6x = -6 is 659, and the absolute value of the function at x=4x = 4 is 341.

References