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Chapter 4: Problem 19

Use the guidelines of this section to make a complete graph of \(f\). $$f(x)=x^{4}-6 x^{2}$$

Short Answer

Expert verified
Question: Identify the key features and graph the function \(f(x) = x^4 - 6x^2\). Answer: The key features of the function \(f(x) = x^4 - 6x^2\) are: 1. Domain: \((-\infty, \infty)\) 2. Range: \((-\infty, \infty)\) 3. x-intercepts: \(x = 0, x = \pm\sqrt{6}\) 4. y-intercept: at the origin (0,0) 5. Symmetry: Even function (symmetric about the y-axis) 6. Critical Points: Local maximum at x = 0; local minimum at x = \(\pm\sqrt{3}\) 7. Inflection Points: x = \(\pm 1\) 8. Concavity: Concave up when \(-1 < x < 1\); concave down when \(x < -1\) or \(x > 1\) 9. Asymptotes: None Using these key features, the graph of the function can be accurately drawn by connecting the critical points and inflection points with the appropriate concavity and considering its symmetry about the y-axis.

Step by step solution

01

Domain and Range

The given function is a polynomial function, and polynomial functions are continuous and defined for all real numbers. Therefore, the domain of \(f(x)\) is all real numbers or \((-\infty, \infty)\). Since the function is a quartic function (highest exponent is 4), the range will be all real numbers as well. Thus, the domain and range of \(f(x)\) are both \((-\infty, \infty)\).
02

Intercepts

To find x-intercepts, set \(f(x) = 0\) and solve for x: $$x^4 - 6x^2 = 0$$ $$x^2(x^2 - 6) = 0$$ $$x^2 = 0, x^2 = 6$$ Now we have 3 x-intercepts: \(x = 0, x = \pm\sqrt{6}\). To find the y-intercept, simply substitute \(x=0\) into the function: $$f(0) = 0^4 - 6(0^2) = 0$$ So, the y-intercept is at the origin (0,0).
03

Symmetry

The given function \(f(x) = x^4 - 6x^2\) is an even function since all the exponents of x are even. This means that the graph is symmetric about the y-axis.
04

Critical Points

To find the critical points, take the first derivative of \(f(x)\) and set it equal to zero: $$f'(x) = 4x^3 - 12x$$ $$f'(x) = 4x(x^2 - 3) = 0$$ The critical points are \(x = 0, \pm\sqrt{3}\). To classify these critical points, we can use the second derivative test. Take the second derivative of \(f(x)\): $$f''(x) = 12x^2 - 12$$ Now, evaluate \(f''(x)\) at the critical points: - \(f''(0) = -12\) (Since this is negative, we have a local maximum at x = 0) - \(f''(\sqrt{3}) = 12\) (Since this is positive, we have a local minimum at x = \(\sqrt{3}\)) - \(f''(-\sqrt{3}) = 12\) (Since this is positive, we have a local minimum at x = \(-\sqrt{3}\))
05

Inflection Points and Concavity

To find inflection points and concavity, analyze the second derivative of \(f(x)\). Set \(f''(x) = 0\): $$12x^2 - 12 = 0$$ $$x^2 = 1$$ $$x = \pm 1$$ These are our inflection points. To determine the concavity of the function, perform the second derivative test: \(f''(x) > 0\), when \(-1 < x < 1\); so, the graph is concave up in this interval. \(f''(x) < 0\), when \(x<-1\) or \(x>1\); so, the graph is concave down in this interval.
06

Asymptotes

Since there is no division by zero and the function is a polynomial, there are no vertical or horizontal asymptotes. Now with all the gathered information, we can make a complete graph of the function \(f(x) = x^4 - 6x^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range of Functions
When dealing with any polynomial function, such as the function given in the exercise, understanding the domain and range is essential. A polynomial function is generally described as continuous, which means it has no breaks, holes, or gaps. It is defined for all real numbers.
Thus, the domain of a polynomial function like \[ f(x) = x^4 - 6x^2 \] is the set of all possible x-values: \((-\infty, \infty)\).
  • The **domain** is all real numbers because you can substitute any real number into the function without causing any mathematical errors, such as division by zero.
  • For the **range**, since it is a polynomial of even degree with a positive leading term, the range is all real numbers and the graph extends indefinitely up and down.
Critical Points
Critical points are significant because they indicate where a function's graph has peaks, valleys, or "turning points." They occur where the derivative of the function equals zero or is undefined. To find the critical points of a function, like \[ f(x) = x^4 - 6x^2 \], follow these steps:
  • Take the **first derivative**: \( f'(x) = 4x^3 - 12x \)
  • Set it equal to zero to find the critical points: \( 4x(x^2 - 3) = 0 \)
The solutions, \( x = 0, \pm\sqrt{3} \), are the critical points. These solutions indicate places on the graph where the slope is zero.
  • **Classification of these points** helps to determine if they represent a local maximum, local minimum, or saddle point, using the second derivative test.
Inflection Points
Inflection points are where the graph of a function changes its concavity, or bends differently. At these points, the second derivative of the function is zero. For \[ f(x) = x^4 - 6x^2 \], the second derivative is \( f''(x) = 12x^2 - 12 \), and setting this equal to zero will give the inflection points:
  • Solve \(12x^2 - 12 = 0\) to find \(x = \pm 1\).
The solutions suggest that the function changes concavity at these points.
  • **Concavity**: When \(f''(x) > 0\), the function is concave up, forming a "U" shape. For \(-1 < x < 1\), the graph is concave up.
  • When \(f''(x) < 0\), the graph is concave down, forming an "n" shape for \(x<-1\) and \(x>1\).
x-intercepts and y-intercepts
Intercepts are helpful for understanding how and where a graph crosses the axes, crucial for sketching graphs.
  • **x-intercepts** are where the graph crosses the x-axis, meaning the output is zero. Set the function equal to zero and solve: \[ x^4 - 6x^2 = 0 \] which gives \(x = 0, \pm\sqrt{6}\).
  • **y-intercept** is where the graph crosses the y-axis, found by setting \(x = 0\). Substituting gives \(f(0) = 0\), hence, the y-intercept is \( (0, 0) \).

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