Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Analyzing a Graph In Exercises \(47-58\), analyze the graph of the function algebraically and use the results to sketch the graph by hand. Then use a graphing utility to confirm your sketch. $$f(x)=\frac{1}{4} x^{4}-2 x^{2}$$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{1}{4} x^{4}-2 x^{2}\) intercepts the x-axis at \(x = 0, -2^{\frac{1}{2}}, 2^{\frac{1}{2}}\) and the y-axis at \(y = 0\). The extrema occur at \(x = 0\) (local maximum) and \(x = -2, 2\) (local minima). As \(x\) approaches \(+\infty\) and \(-\infty\), \(f(x)\) approaches \(+\infty\).

Step 1: Find the Intercepts

To find the x-intercepts (where the graph crosses the x-axis), set \(f(x) = 0\). Solve the equation \(\frac{1}{4} x^{4}-2 x^{2} = 0\). This gives \(x = 0, -2^\frac{1}{2}, 2^\frac{1}{2}\) as the x-intercepts. To find the y-intercept (where the graph crosses the y-axis), plug in \(x = 0\) into the function to get \(f(0) = 0\). So, the y-intercept is \(0\).

Step 2: Calculate the Extrema

Extrema (maximum and minimum points) can be found by setting the first derivative equal to zero and solving for \(x\). The first derivative is \(f'(x) = x^3 - 4x\). Setting this equal to zero gets \(x = 0, -2, 2\). Next, calculate the second derivative \(f''(x) = 3x^2 - 4\) and use it to determine whether these points are maxima, minima, or points of inflection. \(f''(0) = -4 < 0\), so \(x = 0\) is a local maximum. \(f''(-2) = 8 > 0\) and \(f''(2) = 8 > 0\), so \(x = -2\) and \(x = 2\) are local minima.

Step 3: Analyze the End Behavior

The end behavior describes what happens to \(f(x)\) as \(x\) approaches positive or negative infinity. As a fourth degree polynomial, this function has the same behavior at the ends; as \(x\) approaches \(+\infty\) and \(-\infty\), \(f(x)\) approaches \(+\infty\).

Step 4: Sketch the Graph

Plot the intercepts, the extrema, and use the end behavior to sketch a possible graph of the function. Then confirm this sketch using a graphing utility.

Key Concepts

X-Intercepts Calculation

Understanding how to find the x-intercepts of a function is an essential skill in graph analysis. The x-intercepts are the points where a graph crosses the x-axis, which indicate the values of x for which the function equals zero. To calculate these points for a polynomial like the given function, \(f(x)=\frac{1}{4} x^{4}-2 x^{2}\), you'll need to set the equation equal to zero and solve for x.

As demonstrated in the textbook exercise, when we set \(f(x)\) equal to zero, we get \(\frac{1}{4} x^{4}-2 x^{2} = 0\). Factoring this equation can help to simplify the problem. You can pull out an \(x^{2}\) term to get \(x^{2}(\frac{1}{4} x^{2} - 2) = 0\), which tells us that one x-intercept is at \(x = 0\). Further solving \(\frac{1}{4} x^{2} - 2 = 0\) gives two more x-intercepts at \(x = \pm 2^\frac{1}{2}\).

This initial step is crucial because it gives us a good starting point for sketching the graph since the x-intercepts are specific points we know the curve will pass through.

Extrema of Functions

Extrema refer to the maximum and minimum points on a graph, which represent the peaks and valleys of the curve. Calculating the extrema of functions often involves calculus, where the first derivative of the function \(f'(x)\) is used to determine where the slope of the tangent to the curve is zero. In other words, these are the points where the function stops increasing or decreasing and begins to move in the opposite direction.

For the polynomial \(f(x)=\frac{1}{4} x^{4}-2 x^{2}\), we find the first derivative \(f'(x) = x^3 - 4x\). Setting \(f'(x)\) to zero, as done in the example, gives us potential extrema at \(x = 0, -2, 2\). A further check with the second derivative \(f''(x)\), which tells us about the concavity of the function, can confirm the nature of these extrema. The sign of \(f''(x)\) at these points determines whether they are maximums (concave down) or minimums (concave up). In this case, \(f''(0)\) being negative indicates a local maximum at \(x = 0\), and since \(f''(-2)\) and \(f''(2)\) are positive, we know that \(x = -2\) and \(x = 2\) are local minima.

Identifying these critical points allows us to accurately depict the rises and falls of the graph, helping to visualize the function's behavior between and beyond the x-intercepts.

End Behavior of Polynomials

Determining the end behavior of a polynomial function is like predicting the future of a curve. For polynomials, this projective analysis focuses on what happens as the x-values reach extremely large positive or negative numbers. The degree and the leading coefficient of a polynomial are crucial to this understanding.

The function at hand, \(f(x)=\frac{1}{4} x^{4}-2 x^{2}\), is a fourth-degree polynomial with a positive leading coefficient. This combination tells us that as x approaches positive or negative infinity, the function values will also approach positive infinity. This is due to the highest power term dominating the growth of the polynomial at extreme x-values.

In simpler terms, the ends of the curve will both rise upwards as they move away from the x-y plane. When graphing, this knowledge guides the drawing of the curve beyond the plotted extrema. This step is the finishing touch and can often help check earlier steps if the expected behavior doesn't match the generated sketch. Recall, for even-degree polynomials with positive leading coefficients, the arms of the graph will face the same direction—upward—and for negative coefficients, they would both trend downward.