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Chapter 3: Problem 64

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

Short Answer

Expert verified
The polynomial has one local maximum and one local minimum.

Step by step solution

01

Find the Derivative

To find the local maxima and minima, first compute the first derivative of the function, which is necessary to locate critical points. The polynomial function is \( y = \frac{1}{3} x^7 - 17x^2 + 7 \). The first derivative is \( y' = \frac{d}{dx}(\frac{1}{3} x^7 - 17x^2 + 7) = \frac{7}{3}x^6 - 34x \).
02

Find Critical Points

Set the first derivative equal to zero to find the critical points: \( \frac{7}{3}x^6 - 34x = 0 \). Factor the equation to solve for \( x \): \( x(\frac{7}{3}x^5 - 34) = 0 \). The solutions are \( x = 0 \) and \( x = \sqrt[5]{\frac{34}{\frac{7}{3}}} \).
03

Determine the Nature of Critical Points

Find the second derivative to determine if each critical point is a maximum, minimum, or neither. The second derivative is \( y'' = \frac{d^2}{dx^2}(\frac{1}{3} x^7 - 17x^2 + 7) = 14x^5 - 34 \). Substitute critical points into the second derivative: for \( x = 0 \), \( y''(0) = -34 \) (local maximum); for \( x = \sqrt[5]{\frac{102}{7}} \), \( y'' \) is positive (local minimum).
04

Conclude the Number of Local Maxima and Minima

Based on the second derivative test, the function has one local maximum at \( x = 0 \) and one local minimum at \( x = \sqrt[5]{\frac{34}{\frac{7}{3}}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
When graphing a polynomial and seeking to find its local maxima and minima, the first derivative is your go-to tool. The first derivative of a function, denoted as \( y' \), provides the rate at which the function's value is changing. For our polynomial function \( y = \frac{1}{3} x^7 - 17x^2 + 7 \), the first derivative is calculated as follows:
  • Differentiate the polynomial: \( y' = \frac{7}{3}x^6 - 34x \).
  • This derivative tells us how steep or flat the function is at any point \( x \).
  • A critical step in finding local maxima and minima involves setting this first derivative to zero and solving for \( x \): \( \frac{7}{3} x^6 - 34x = 0 \).
Breaking this down, the solutions to this equation provide the critical points, which are potential locations for local maxima or minima. These points are vital to understanding where the graph changes from increasing to decreasing, or vice versa.
Critical Points
Critical points occur where the first derivative is equal to zero or is undefined, signaling possible peaks and troughs in the graph of a polynomial. These points are where the slope of the curve is neither increasing nor decreasing. For the function given, the critical points are derived by solving:
  • Set the first derivative to zero: \( \frac{7}{3} x^6 - 34x = 0 \).
  • Factor the equation: \( x(\frac{7}{3} x^5 - 34) = 0 \).
  • This gives us potential critical points at \( x = 0 \) and \( x = \sqrt[5]{\frac{34}{\frac{7}{3}}} \).
Critical points are essential in the sketching and understanding of the function's graph as they denote regions where the function can change directions. However, not all critical points are maxima or minima; their nature needs further examination which involves the second derivative.
Second Derivative
The second derivative of a function, denoted as \( y'' \), gives further insight into the curvature of the graph and helps determine the nature of each critical point. It tells us whether a critical point is a local maximum, minimum, or neither.
  • The second derivative is calculated as \( y'' = 14x^5 - 34 \) for our polynomial.
  • By substituting the critical points into this derivative, we can apply the second derivative test.
    • If \( y''(x) > 0 \), the function is concave up at \( x \) and the critical point is a local minimum.
    • If \( y''(x) < 0 \), the function is concave down, indicating a local maximum.
For \( x = 0 \), \( y''(0) = -34 \), showing a local maximum. For \( x = \sqrt[5]{\frac{34}{\frac{7}{3}}} \), since \( y'' \) is positive, this critical point is a local minimum. This second derivative test thus confirms the types of extremas at the identified critical points.

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Most popular questions from this chapter

Graph the rational function and find all vertical asymptotes, \(x\)- and \(y\)-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$r(x)=\frac{x^{4}-3 x^{3}+6}{x-3}$$

Depth of Snowfall Snow began falling at noon on Sunday. The amount of snow on the ground at a certain location at time \(t\) was given by the function $$\begin{aligned} h(t)=11.60 t &-12.41 t^{2}+6.20 t^{3} \\ &-1.58 t^{4}+0.20 t^{5}-0.01 t^{6} \end{aligned}$$ where \(t\) is measured in days from the start of the snowfall and \(h(t)\) is the depth of snow in inches. Draw a graph of this function and use your graph to answer the following questions. (a) What happened shortly after noon on Tuesday? (b) Was there ever more than 5 in. of snow on the ground? If so, on what day(s)? (c) On what day and at what time (to the nearest hour) did the snow disappear completely?

How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to find such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) A polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefficients if it has no real zeros?

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$r(x)=\frac{2 x^{3}+2 x}{x^{2}-1}$$

Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$P(x)=8 x^{3}+10 x^{2}-39 x+9 ; \quad a=-3, b=2$$
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