Problem 12 - Calculus Early Transcendentals

Question

Sketch a curve with the following properties. $$f(x)=2 x^{6}-3 x^{4}$$

Step-by-Step Solution

Answer

Question: Sketch the curve of the function \(f(x)=2x^6 - 3x^4\) using its critical properties. Answer: The curve of the function \(f(x)=2x^6 - 3x^4\) is symmetric about the y-axis, has x-intercepts at \(x=0\) and \(x= \pm \sqrt{\frac{3}{2}}\), and y-intercept at \((0,0)\). Its critical points are at \(x=0\), \(x=-1\), and \(x=1\). The function is increasing on \((-\infty, -1)\) and \((1, \infty)\), and decreasing on \((-1, 0)\) and \((0, 1)\).

Step 1: 1. Determine Symmetry

To determine if the function has symmetry, we need to see if it is an even or odd function. An even function satisfies \(f(-x) = f(x)\), while an odd function satisfies \(f(-x) = -f(x)\). Let's check if the function is even or odd: $$f(-x) = 2(-x)^6 - 3(-x)^4 = 2x^6 - 3x^4 = f(x)$$ Since \(f(-x) = f(x)\), the function is even and has symmetry about the y-axis.

Step 2: 2. Find Intercepts

To find the x-intercepts of the function, set \(f(x) = 0\) and solve for x: $$2x^6 - 3x^4 = 0$$ Factor out \(x^4\), we get: $$x^4(2x^2 - 3) = 0$$ This expression equals zero when \(x^4 = 0\) or \(2x^2 - 3 = 0\). We get the following x-intercepts: \(x = 0\) and \(x = \pm \sqrt{\frac{3}{2}}\) For the y-intercept, we just need to find the value of the function at \(x=0\): $$f(0) = 2(0)^6 - 3(0)^4 = 0$$ Therefore, the y-intercept is at the origin, \((0, 0)\).

Step 3: 3. Find Critical Points

To find the critical points of the function, we need to take the first derivative and set it equal to zero. Then, we will find the points where the function changes its increasing/decreasing behavior. $$f'(x) = 12x^5 - 12x^3$$ Set \(f'(x) = 0\), we get: $$12x^5 - 12x^3 = 0$$ Factor the equation, we have: $$12x^3(x^2-1) = 0$$ The critical points are \(x = 0\), \(x = -1\), and \(x = 1\).

Step 4: 4. Determine Intervals of Increase/Decrease

To determine the intervals of increase or decrease, we will test the intervals surrounding the critical points in the first derivative. Let's pick numbers in each interval and compute the value of \(f'(x)\). For \(x<-1\): Choose \(x=-2\), we have \(f'(-2) > 0\). Therefore, the function is increasing on \((-\infty, -1)\). For \(-11\): Choose \(x=2\), we have \(f'(2) > 0\). Therefore, the function is increasing on \((1, \infty)\). Now we can sketch the curve using all these properties found in steps 1-4: - The curve is symmetric about the y-axis. - It has x-intercepts at \(x=0\) and \(x= \pm \sqrt{\frac{3}{2}}\), and y-intercept at \((0,0)\). - There are critical points at \(x=0\), \(x=-1\), and \(x=1\). - The function is increasing on \((-\infty, -1)\) and \((1, \infty)\), and decreasing on \((-1, 0)\) and \((0, 1)\). Using these properties, you can now draw a sketch of the curve for the function \(f(x)=2x^6 - 3x^4\).

Key Concepts

Function Symmetry

In mathematics, understanding function symmetry is crucial. It helps in simplifying the analysis of functions. A function is symmetric around the y-axis if it is an even function.

• An even function satisfies the condition: \( f(-x) = f(x) \).
• An example is the function \( f(x) = 2x^6 - 3x^4 \). For this function, plugging \(-x\) yields the same expression as \( f(x) \).
• This means the graph of the function is mirror-symmetrical around the y-axis.

Even functions are visually balanced. This symmetry allows one to predict the shape of the graph on one side of the y-axis if you know the other.

Critical Points

Critical points of a function are spots where the graph changes its direction. They are found by setting the derivative of the function to zero.

• The first derivative identifies these points. For \( f(x) = 2x^6 - 3x^4 \), the first derivative is \( f'(x) = 12x^5 - 12x^3 \).
• Set \( f'(x) = 0 \), giving us the equation \( 12x^3( x^2 - 1 ) = 0 \).
• Solve it to find the critical points: \( x = 0 \), \( x = -1 \), and \( x = 1 \).

These points are where potential local maxima or minima occur. It is important to analyze these points further to understand the intervals of increase or decrease.

Intervals of Increase/Decrease

The intervals in which a function is increasing or decreasing show how the graph moves. Analyzing these intervals involves using the first derivative to test the critical points.

• Choose test values from intervals around critical points and substitute them into \( f'(x) \).
• For \( x < -1 \), choose \( x = -2 \), leading to \( f'(-2) > 0 \), indicating an increase.
• For \( -1 < x < 0 \), choose \( x = -\frac{1}{2} \), yielding \( f'(-\frac{1}{2}) < 0 \), showing a decrease.
• For \( 0 < x < 1 \), choose \( x = \frac{1}{2} \), resulting in \( f'(\frac{1}{2}) < 0 \), another decrease.
• For \( x > 1 \), choose \( x = 2 \), giving \( f'(2) > 0 \), pointing to an increase.

These intervals help paint the bigger picture of the function's behavior, guiding us to understand the nature of the graph.

Polynomial Functions

Polynomial functions are expressions formed by variables raised to whole number powers and combined with coefficients. Being one of the simplest forms of functions, they are key in algebra.

• The function \( f(x) = 2x^6 - 3x^4 \) is a polynomial of degree six. It contains even powers of \( x \), which contributes to its even symmetry.
• Polynomials like this have a clear, defined structure, often leading to polynomial equations being easy to factor.
• The graph of a polynomial function is smooth and continuous, with no breaks or holes, making them easy to sketch.

Understanding polynomial functions is crucial for deeper mathematical analysis, providing a strong base for calculus and other higher-level math topics.