Problem 58 - Calculus Early Transcendentals

Question

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=4 x^{3}-9 x^{4} $$

Step-by-Step Solution

Answer

The intercepts are (0,0) and (4/9,0); stationary points at (0,0) and (1/3, 1/27); inflection point at (2/9, 8/729).

Step 1: Find the Intercepts

To find the intercepts, we identify where the polynomial crosses the axes.For the **y-intercept**, set \(x = 0\):\[p(0) = 4(0)^3 - 9(0)^4 = 0\]The y-intercept is at \((0,0)\).For the **x-intercepts**, set \(p(x) = 0\):\[4x^3 - 9x^4 = 0 \x^3(4 - 9x) = 0\]This gives us solutions at \(x = 0\) and \(x = \frac{4}{9}\). Thus, the x-intercepts are \((0,0)\) and \(\left(\frac{4}{9},0\right)\).

Step 2: Find the Stationary Points

Stationary points occur where the first derivative \(p'(x)\) equals zero. First, calculate \(p'(x)\):\[p(x) = 4x^3 - 9x^4 \p'(x) = 12x^2 - 36x^3\]Set the derivative to zero and solve:\[12x^2 - 36x^3 = 0 \12x^2(1 - 3x) = 0\]This gives \(x = 0\) and \(x = \frac{1}{3}\) as solutions.Substitute into \(p(x)\) to find coordinates:- \(p(0) = 0\), stationary point at \((0,0)\).- \(p\left(\frac{1}{3}\right) = 4\left(\frac{1}{3}\right)^3 - 9\left(\frac{1}{3}\right)^4 = \frac{4}{27} - \frac{9}{81} = \frac{1}{27}\), stationary point at \(\left(\frac{1}{3}, \frac{1}{27}\right)\).

Step 3: Determine Inflection Points

Inflection points occur where the second derivative \(p''(x)\) changes sign. First, calculate \(p''(x)\):\[p'(x) = 12x^2 - 36x^3 \p''(x) = 24x - 108x^2\]Set \(p''(x) = 0\) and solve:\[24x - 108x^2 = 0 \24x(1 - 4.5x) = 0\]This gives \(x = 0\) and \(x = \frac{2}{9}\) as potential inflection points. Check the sign change:- Test intervals around \(0\) and \(\frac{2}{9}\).- Confirm: \(x = 0\) is not an inflection point due to no sign change.- \(x = \frac{2}{9}\) results in a sign change across \(p''(x)\), therefore an inflection point.Substitute to find the inflection point coordinates:- \(p\left(\frac{2}{9}\right) = 4\left(\frac{2}{9}\right)^3 - 9\left(\frac{2}{9}\right)^4 = \frac{32}{729} - \frac{144}{6561} = \frac{8}{729}\), inflection point at \(\left(\frac{2}{9}, \frac{8}{729}\right)\).

Step 4: Confirm with a Graphing Utility

Use a graphing utility to plot \(p(x) = 4x^3 - 9x^4\). Check that the graph intersects at the intercepts found: \((0,0)\) and \(\left( \frac{4}{9}, 0 \right)\). Verify stationary points \((0,0)\) and \(\left( \frac{1}{3}, \frac{1}{27} \right)\), as well as the inflection point \(\left( \frac{2}{9}, \frac{8}{729} \right)\). The graph should confirm the calculations with correct symmetry and curvature changes according to the problem's polynomial properties.

Key Concepts

Polynomial Intercepts

When we're talking about polynomial intercepts, we're referring to the points where the graph of the polynomial crosses the axes. The two main types of intercepts are the x-intercept and the y-intercept.

• **Y-intercept:** This is the point where the graph crosses the y-axis. To find it, we set all the x-values to zero. For the function \(p(x) = 4x^3 - 9x^4\), when \(x = 0\), the polynomial evaluates to zero: \(p(0) = 0\). So, the y-intercept is at \((0,0)\).


• **X-intercept:** These points are found by setting the polynomial equal to zero and solving for x. For \(p(x) = 4x^3 - 9x^4\), factor the expression: \(x^3(4 - 9x) = 0\). This gives solutions \(x = 0\) and \(x = \frac{4}{9}\). Thus, x-intercepts are at \((0,0)\) and \((\frac{4}{9}, 0)\).

Intercepts are crucial because they provide key reference points for sketching the graph of the polynomial.

Stationary Points

Stationary points on a graph occur where the slope is zero, meaning the graph has a horizontal tangent line. For polynomials, these points are found using the first derivative. To find the stationary points for \(p(x) = 4x^3 - 9x^4\), calculate the first derivative: \(p'(x) = 12x^2 - 36x^3\). Set \(p'(x) = 0\) to find: \[12x^2(1 - 3x) = 0\] Solving gives us \(x = 0\) and \(x = \frac{1}{3}\).

• For \(x = 0\): Plug back into the original polynomial \(p(0) = 0\), indicating a stationary point at \((0,0)\).
• For \(x = \frac{1}{3}\): Calculate \(p(\frac{1}{3}) = \frac{4}{27} - \frac{9}{81} = \frac{1}{27}\), giving another stationary point at \((\frac{1}{3}, \frac{1}{27})\).

These stationary points help to determine sections of the graph where the polynomial changes from increasing to decreasing and vice versa.

Inflection Points in Calculus

Inflection points are where the curve changes concavity, meaning it switches from being "cup-up" to "cup-down" or vice versa. These are found using the second derivative.To find the inflection points for \(p(x) = 4x^3 - 9x^4\), we first find the second derivative: \(p''(x) = 24x - 108x^2\). Setting it to zero, we get:\[24x(1 - 4.5x) = 0\]Finding solutions gives \(x = 0\) and \(x = \frac{2}{9}\) as potential inflection points.

• Test around \(x = 0\): Observing \(p''(x)\), there is no sign change, confirming no inflection at \((0,0)\).
• Test around \(x = \frac{2}{9}\): A sign change confirms a true inflection point. Calculate \(p(\frac{2}{9}) = \frac{8}{729}\), so the inflection point is at \((\frac{2}{9}, \frac{8}{729})\).

Understanding inflection points is important as they mark where the graph's curvature changes, impacting the overall shape of the polynomial's graph.