Problem 80
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
Verified Answer
P(x)=x^3-4x has 3 x-intercepts, 2 extrema; Q(x)=x^3+4x has 1 x-intercept, no extrema. Generally, P(x)=x^3-ax has 3 x-intercepts, 2 extrema; Q(x)=x^3+ax has 1 x-intercept, no extrema for any positive a.
1Step 1: Find x-intercepts for P(x)
The polynomial is given by \( P(x) = x^3 - 4x \). To find the x-intercepts, set \( P(x) = 0 \):\[ x^3 - 4x = 0 \]Factor the equation:\[ x(x^2 - 4) = 0 \]Set each factor to zero:* \( x = 0 \)* \( x^2 - 4 = 0 \)Solve \( x^2 - 4 = 0 \):\[ x^2 = 4 \]\[ x = \pm 2 \]The x-intercepts are \( x = 0, 2, -2 \), totaling 3 x-intercepts.
2Step 2: Find local extrema for P(x)
To find the local extrema, calculate the derivative of \( P(x) \):\[ \frac{d}{dx}(x^3 - 4x) = 3x^2 - 4 \]Set the derivative to zero to find critical points:\[ 3x^2 - 4 = 0 \]Solve for \( x \):\[ 3x^2 = 4 \]\[ x^2 = \frac{4}{3} \]\[ x = \pm \sqrt{\frac{4}{3}} \]There are two critical points, thus potentially two local extrema.
3Step 3: Find x-intercepts for Q(x)
The polynomial is \( Q(x) = x^3 + 4x \).Set \( Q(x) = 0 \) to find x-intercepts:\[ x^3 + 4x = 0 \]Factor the equation:\[ x(x^2 + 4) = 0 \]Set each factor to zero:* \( x = 0 \)* \( x^2 + 4 = 0 \)The equation \( x^2 + 4 = 0 \) has no real solutions. Thus, there is 1 x-intercept at \( x = 0 \).
4Step 4: Find local extrema for Q(x)
Take the derivative of \( Q(x) \):\[ \frac{d}{dx}(x^3 + 4x) = 3x^2 + 4 \]Set the derivative equal to zero to find critical points:\[ 3x^2 + 4 = 0 \]This equation has no real roots since \( 3x^2 = -4 \) is impossible over the reals. Therefore, there are no local extrema for \( Q(x) \).
5Step 5: Analyze general case for P(x)=x^3-ax and Q(x)=x^3+ax
For \( P(x) = x^3 - ax \), set \( P(x) = 0 \):\[ x(x^2 - a) = 0 \]x-intercepts occur at \( x = 0 \) and \( x = \pm \sqrt{a} \), providing 3 x-intercepts. The derivative \( P'(x) = 3x^2 - a \) leads to two critical points at \( x = \pm \sqrt{a/3} \), allowing for 2 local extrema.For \( Q(x) = x^3 + ax \), set \( Q(x) = 0 \):\[ x(x^2 + a) = 0 \]Gives one x-intercept at \( x = 0 \) as \( x^2 + a = 0 \) implies no real roots. The derivative \( Q'(x) = 3x^2 + a \), having no solutions for \( x \) over the reals, produces no local extrema.
Key Concepts
Understanding X-Intercepts in Polynomial FunctionsIdentifying Local Extrema in PolynomialsExploring Critical Points in Polynomial AnalysisDistinguishing Real Roots in Polynomials
Understanding X-Intercepts in Polynomial Functions
Polynomial functions often intersect the x-axis at specific values known as x-intercepts. These intercepts are found by solving the equation when the polynomial is set to zero. For example, consider the polynomial \( P(x) = x^3 - 4x \). To find its x-intercepts, set the polynomial equal to zero: \( x^3 - 4x = 0 \). Next, factor the equation to make it easier to solve: \( x(x^2 - 4) = 0 \).
Now, set each factor equal to zero: \(x = 0\) or \(x^2 - 4 = 0\). Solving \(x^2 - 4 = 0\) gives \(x = \pm 2\). Thus, the x-intercepts are \(x = 0, 2, -2\), indicating three points where the graph touches the x-axis.
In contrast, for \(Q(x) = x^3 + 4x\), the equation \(x(x^2 + 4) = 0\) results in an x-intercept only at \(x = 0\) since \(x^2 + 4 = 0\) provides no real roots. Understanding these intercepts is pivotal when analyzing the graph of polynomial functions.
Now, set each factor equal to zero: \(x = 0\) or \(x^2 - 4 = 0\). Solving \(x^2 - 4 = 0\) gives \(x = \pm 2\). Thus, the x-intercepts are \(x = 0, 2, -2\), indicating three points where the graph touches the x-axis.
In contrast, for \(Q(x) = x^3 + 4x\), the equation \(x(x^2 + 4) = 0\) results in an x-intercept only at \(x = 0\) since \(x^2 + 4 = 0\) provides no real roots. Understanding these intercepts is pivotal when analyzing the graph of polynomial functions.
Identifying Local Extrema in Polynomials
Local extrema in a polynomial function are the points where the function reaches a local maximum or minimum. To find these points, the derivative of the polynomial is determined and set to zero to solve for critical points.
For \( P(x) = x^3 - 4x \), the derivative is \( P'(x) = 3x^2 - 4 \). Setting \( P'(x) = 0 \) results in \( 3x^2 = 4 \), or \( x^2 = \frac{4}{3} \), giving critical points at \( x = \pm \sqrt{\frac{4}{3}} \). These critical points indicate up to two local extrema, which are specific x-values where the graph's slope changes from increasing to decreasing, or vice versa.
On the other hand, for \( Q(x) = x^3 + 4x \), solving the derivative \( Q'(x) = 3x^2 + 4 = 0 \) shows that there are no real roots, implying no critical points and thus no local extrema. Identifying local extrema is essential for understanding the shape and behavior of polynomial graphs.
For \( P(x) = x^3 - 4x \), the derivative is \( P'(x) = 3x^2 - 4 \). Setting \( P'(x) = 0 \) results in \( 3x^2 = 4 \), or \( x^2 = \frac{4}{3} \), giving critical points at \( x = \pm \sqrt{\frac{4}{3}} \). These critical points indicate up to two local extrema, which are specific x-values where the graph's slope changes from increasing to decreasing, or vice versa.
On the other hand, for \( Q(x) = x^3 + 4x \), solving the derivative \( Q'(x) = 3x^2 + 4 = 0 \) shows that there are no real roots, implying no critical points and thus no local extrema. Identifying local extrema is essential for understanding the shape and behavior of polynomial graphs.
Exploring Critical Points in Polynomial Analysis
Critical points in a polynomial function are where the function's derivative equals zero or is undefined. These points often coincide with local extrema but can exist independently when no extrema are present.
In the function \( P(x) \), derived as \( P'(x) = 3x^2 - 4 \), setting the derivative to zero, \( 3x^2 - 4 = 0 \), finds potential local maxima or minima at \( x = \pm \sqrt{\frac{4}{3}} \).
For \( Q(x) \), the derivative \( Q'(x) = 3x^2 + 4 \) equals zero has no real solutions, indicating the absence of critical points. The lack of solutions is due to the irreducibility of \( 3x^2 + 4 = 0 \) with real numbers.
Critical points are vital when sketching the behavior of polynomial functions as they determine where the function changes slope, informing our understanding of the function's curvature and points of interest.
In the function \( P(x) \), derived as \( P'(x) = 3x^2 - 4 \), setting the derivative to zero, \( 3x^2 - 4 = 0 \), finds potential local maxima or minima at \( x = \pm \sqrt{\frac{4}{3}} \).
For \( Q(x) \), the derivative \( Q'(x) = 3x^2 + 4 \) equals zero has no real solutions, indicating the absence of critical points. The lack of solutions is due to the irreducibility of \( 3x^2 + 4 = 0 \) with real numbers.
Critical points are vital when sketching the behavior of polynomial functions as they determine where the function changes slope, informing our understanding of the function's curvature and points of interest.
Distinguishing Real Roots in Polynomials
Real roots are solutions to the polynomial equation where the polynomial equals zero, and the solutions are real numbers. These roots correspond directly to the x-intercepts of the function.
Consider \( P(x) = x^3 - ax \). Setting this zero results in \( x(x^2 - a) = 0 \), producing real roots at \( x = 0 \) and \( x = \pm \sqrt{a} \). This means that for \( P(x) \) when \( a > 0 \), three real roots or zeroes occur.
For \( Q(x) = x^3 + ax \), solving zero gives \( x(x^2 + a) = 0 \). Here, real roots are seen only at \( x = 0 \) since \( x^2 + a = 0 \) has no real solutions, depending on the sign and value of \( a \).
Understanding real roots, synonymous with x-intercepts, plays a crucial role in graphing polynomials as they show where the function crosses or touches the x-axis, offering a clear visual representation of the function's solutions.
Consider \( P(x) = x^3 - ax \). Setting this zero results in \( x(x^2 - a) = 0 \), producing real roots at \( x = 0 \) and \( x = \pm \sqrt{a} \). This means that for \( P(x) \) when \( a > 0 \), three real roots or zeroes occur.
For \( Q(x) = x^3 + ax \), solving zero gives \( x(x^2 + a) = 0 \). Here, real roots are seen only at \( x = 0 \) since \( x^2 + a = 0 \) has no real solutions, depending on the sign and value of \( a \).
Understanding real roots, synonymous with x-intercepts, plays a crucial role in graphing polynomials as they show where the function crosses or touches the x-axis, offering a clear visual representation of the function's solutions.
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