Problem 80 - Precalculus Mathematics for Calculus

Question

(a) How many \(x\) -intercepts and how many local extrema does the polynomial \(P(x)=x^{3}-4 x\) have? (b) How many \(x\) -intercepts and how many local extrema does the polynomial \(Q(x)=x^{3}+4 x\) have? (c) If \(a>0,\) how many \(x\) -intercepts and how many local extrema does each of the polynomials \(P(x)=x^{3}-a x\) and \(Q(x)=x^{3}+a x\) have? Explain your answer.

Step-by-Step Solution

Answer

(a) P(x): 3 intercepts, 2 extrema. (b) Q(x): 1 intercept, no extrema. (c) P(x): 3 intercepts, 2 extrema; Q(x): 1 intercept, no extrema.

Step 1: Finding x-intercepts of P(x)

To find the x-intercepts of the polynomial \(P(x)=x^3-4x\), we set \(P(x) = 0\):\[x^3 - 4x = 0\]Factor out \(x\):\[x(x^2 - 4) = 0\]Then solve for \(x\):- \(x = 0\)- \(x^2 - 4 = 0\) which gives \(x = \pm 2\)Thus, the polynomial \(P(x)\) has 3 x-intercepts: \(x = -2, 0, 2\).

Step 2: Finding local extrema of P(x)

To find the local extrema, first find the derivative of \(P(x) = x^3 - 4x\):\[P'(x) = 3x^2 - 4\]Set the derivative equal to zero to locate critical points:\[3x^2 - 4 = 0\]\[3x^2 = 4\]\[x^2 = \frac{4}{3}\]\[x = \pm \sqrt{\frac{4}{3}}\]These critical points indicate two local extrema.

Step 3: Finding x-intercepts of Q(x)

Solve \(Q(x) = x^3 + 4x = 0\) to find x-intercepts:\[x(x^2+4) = 0\]Solve for \(x\):- \(x = 0\)- \(x^2 + 4 = 0\) has no real roots (as \(x^2 = -4\)).Thus, \(Q(x)\) has one x-intercept: \(x = 0\).

Step 4: Finding local extrema of Q(x)

Find the derivative of \(Q(x) = x^3 + 4x\):\[Q'(x) = 3x^2 + 4\]Since \(3x^2 + 4 eq 0\) for any real \(x\), the polynomial has no critical points and hence no local extrema.

Step 5: General case for P(x) and Q(x)

For \(P(x) = x^3 - ax\), solve \[x^3 - ax = 0\] by factoring:\[x(x^2 - a) = 0\]This results in roots \(x = 0, \pm \sqrt{a}\), so there are 3 intercepts.Next, find the derivative: \[P'(x) = 3x^2 - a\]Set to zero for critical points: \[x = \pm \sqrt{\frac{a}{3}}\], indicating 2 local extrema.For \(Q(x) = x^3 + ax\), solve \[x^3 + ax = 0\] by factoring:\[x(x^2 + a) = 0\]This gives \(x = 0\) as the only real root, resulting in 1 intercept.The derived equation \(Q'(x) = 3x^2 + a\) has no real solutions for critical points, hence no local extrema.

Key Concepts

X-intercepts

The x-intercepts of a polynomial are the points where the graph of the polynomial crosses the x-axis. These points occur when the polynomial's value is zero. To find the x-intercepts of the polynomial, set the polynomial equation equal to zero and solve for the values of x. Let's take a look at some examples:

• For the polynomial \(P(x) = x^3 - 4x\), setting \(P(x) = 0\) leads to \(x(x^2 - 4) = 0\). This can be factored to find the x-values: \(x = 0\), \(x = 2\) and \(x = -2\). Thus, \(P(x)\) has three x-intercepts.
• In the case of \(Q(x) = x^3 + 4x\), solving \(Q(x) = 0\) gives \(x(x^2 + 4) = 0\). Here, \(x = 0\) is the only real solution since \(x^2 + 4\) doesn’t yield real roots, resulting in one x-intercept.

When accounting for a positive constant \(a\), the polynomials \(P(x) = x^3 - ax\) and \(Q(x) = x^3 + ax\) are considered. Solving \(x^3 - ax = 0\) and \(x^3 + ax = 0\) again reveals three and one x-intercepts, respectively. Recognizing these intercepts helps visualize where the polynomial hits the x-axis, enabling predictions about the behavior of the graph.

Local Extrema

Local extrema refer to the highest or lowest points in a specific interval on the graph of a function. They are the points where the function changes direction from increasing to decreasing or vice versa. These are identified by finding the derivative of the polynomial and setting it equal to zero to determine critical points.

• For \(P(x) = x^3 - 4x\), the derivative is \(P'(x) = 3x^2 - 4\). Setting \(3x^2 - 4 = 0\) results in the critical points \(x = \pm \sqrt{\frac{4}{3}}\). These critical points indicate two local extrema because there is a change in the sign of the derivative—a concept central to the first derivative test.
• For \(Q(x) = x^3 + 4x\), the derivative \(Q'(x) = 3x^2 + 4\) does not equal zero for any real \(x\), implying there are no critical points and, therefore, no local extrema.

Local extrema are important for understanding the shape of the polynomial graph, as they represent peaks and valleys. They are particularly useful when graphing the function, as they help indicate where the graph curves.

Critical Points

Critical points of a polynomial occur where the derivative is zero or does not exist. These points help identify possible locations of local extrema, which are the troughs and peaks in the graph of the function.

• For \(P(x) = x^3 - 4x\), the critical points are found by computing the derivative \(P'(x) = 3x^2 - 4\). Solving \(3x^2 - 4 = 0\) yields \(x = \pm \sqrt{\frac{4}{3}}\), marking potential points for local extrema.
• For the polynomial \(Q(x) = x^3 + 4x\), no real critical points exist as the derivative \(Q'(x) = 3x^2 + 4\) cannot be zero for any real \(x\).

Critical points essentially serve as a guide for examining the behavior of the polynomial's curve. They indicate where the slope of the curve changes and help determine where the function increases or decreases. Identifying these points is crucial in the analysis of functions, especially in calculus, to provide insights into the graph's comprehensive layout.