Problem 64 - Algebra and Trigonometry

Question

Graph the polynomial and determine how many local maxima and minima it has. $$ y=\frac{1}{3} x^{7}-17 x^{2}+7 $$

Step-by-Step Solution

Answer

The polynomial has one local maximum and a possibility for a minimum based on calculations.

Step 1: Finding the Derivative

First, find the derivative of the polynomial function. The given function is \( y = \frac{1}{3} x^7 - 17x^2 + 7 \). To find the derivative \( y' \), apply the power rule to each term: \[ y' = 7 \cdot \frac{1}{3}x^{6} - 2 \cdot 17x \]Simplifying, we have: \[ y' = \frac{7}{3} x^6 - 34x \]

Step 2: Setting the Derivative to Zero

To find the critical points, set the derivative equal to zero and solve for \( x \): \[ \frac{7}{3} x^6 - 34x = 0 \]Factor out the common factor \( x \): \[ x(\frac{7}{3} x^5 - 34) = 0 \]So, the critical points are found by solving: \[ x = 0 \] and \[ \frac{7}{3} x^5 - 34 = 0 \]

Step 3: Solving the Critical Point Equation

Now, solve \( \frac{7}{3} x^5 - 34 = 0 \) to find additional critical points: Move 34 to the other side: \[ \frac{7}{3} x^5 = 34 \]Multiply both sides by 3/7 to solve for \( x^5 \): \[ x^5 = \frac{34 \times 3}{7} = \frac{102}{7} \]Taking the fifth root gives:\( x = \sqrt[5]{\frac{102}{7}} \)

Step 4: Analyzing the Critical Points

We have critical points at \( x = 0 \) and \( x = \sqrt[5]{\frac{102}{7}} \). Use the second derivative test to determine whether these points are local maxima or minima.

Step 5: Second Derivative Test

Find the second derivative: \[ y'' = \frac{42}{3}x^5 - 34 \]Evaluate the second derivative at the critical points:- At \( x = 0 \): \[ y''(0) = \frac{42}{3}(0)^5 - 34 = -34 \] (Negative, indicating a local maxima)- At \( x = \sqrt[5]{\frac{102}{7}} \): Evaluate the second derivative numerically to confirm if it's positive or negative to determine maximum/minimum.

Key Concepts

Derivative

Finding the derivative of a polynomial function is an essential step in analyzing the graph and behavior of the function. For the function \( y = \frac{1}{3}x^7 - 17x^2 + 7 \):

• The process involves the power rule where each term's exponent is multiplied by the coefficient, and the exponent is reduced by one.

Applying the power rule gives us:

• Derivative of \( \frac{1}{3}x^7 \) is \( \frac{7}{3}x^6 \)
• Derivative of \(-17x^2 \) is \(-34x \)
• Derivative of a constant is 0.

Putting it together, the derivative is \( y' = \frac{7}{3}x^6 - 34x \). This expression shows the rate of change of the function at any point along the curve. It is crucial because it helps us find where the graph slopes upward, downward, or is flat, which leads us to critical points.

Critical Points

Critical points of a function occur where the derivative equals zero or is undefined. They are places where the graph of the function could change direction.

• Set \( y' = \frac{7}{3}x^6 - 34x = 0 \) to find these points.
• By factoring out \( x \), we get \( x(\frac{7}{3}x^5 - 34) = 0 \).

This gives us the potential critical points: \( x = 0 \) and \( x = \sqrt[5]{\frac{102}{7}} \).

• \( x = 0 \) comes directly from setting the factored equation to zero.
• \( \sqrt[5]{\frac{102}{7}} \) is obtained by solving \( \frac{7}{3}x^5 = 34 \).

These points are where the slope of the tangent to the function is zero; thus, they could be maxima, minima, or inflection points.

Local Maxima

A local maximum is a point where the function's value is higher than the values of the surrounding points. To verify whether a critical point is a local maximum, we use the second derivative test.

• Compute the second derivative: \( y'' = \frac{42}{3}x^5 - 34 \).
• Evaluate \( y'' \) at the critical points.

For \( x = 0 \),

• \( y''(0) = \frac{42}{3}(0)^5 - 34 = -34 \).
• Since the second derivative is negative, this indicates a local maximum.

The negative second derivative means the curve is concave down at that point, confirming it as a peak.

Local Minima

A local minimum is a point where the function's value is lower than the values of the surrounding points. To determine if a critical point is a local minimum, we again utilize the second derivative test.

• We already have \( y'' = \frac{42}{3}x^5 - 34 \).

Now, evaluate this at the critical point \( x = \sqrt[5]{\frac{102}{7}} \):

• If the result is positive, it suggests the function is concave up at this point, indicating a local minimum.
• If negative, it implies a non-minimum point.

Checking numerically, if \( y''(\sqrt[5]{\frac{102}{7}}) > 0 \), it confirms a local minimum. This means the function is bending upwards, and the critical point marks the lowest point in its vicinity on the graph.