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Chapter 3: Problem 15

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{4}-4 x^{3}+6 $$

Short Answer

Expert verified
a) The function f(x) is decreasing on \((-\infty, 0) \cup (0, 3)\) and increasing on \((3, \infty)\). b) There is a relative minimum at the point \((3, -21)\).

Step by step solution

01

Find the first derivative of the function

To find the first derivative, we will differentiate the function f(x) with respect to x. \(f'(x) = \frac{d}{dx} (x^4 -4x^3 + 6)\) Using the power rule, \(f'(x) = 4x^3 - 12x^2\)
02

Find the critical points

Critical points occur when the first derivative is equal to 0 or f'(x) is undefined. In our case, the derivative is a polynomial, so there is no x value where f'(x) is undefined. Now, we will find the roots of f'(x) by setting it equal to 0. \(4x^3 - 12x^2 = 0\) Factor out the common term \(4x^2\) \(4x^2(x - 3) = 0\) Now, we find the roots: \(x_1 = 0\) \(x_2 = 3\) So, we have two critical points at \(x = 0\) and \(x = 3\).
03

Determine where f'(x) is positive or negative

Now let's analyze the intervals in which \(f'(x)\) is positive or negative. This will tell us where the function is increasing or decreasing. Test each interval with a point to see the sign of \(f'(x)\): Between \(-\infty\) and \(0\), choose any point, like \(x = -1\): \(f'(-1) = 4(-1)^3 - 12(-1)^2 = -4 - 12 < 0\) Between \(0\) and \(3\), choose any point, like \(x = 1\): \(f'(1) = 4(1)^3 - 12(1)^2 = 4 - 12 < 0\) Between \(3\) and \(\infty\), choose any point, like \(x = 4\) \(f'(4) = 4(4)^3 - 12(4)^2 = 256 - 192 > 0\)
04

Identify the intervals and relative extremas

We found that: - \(f'(x) < 0\) for \(-\infty < x < 3\), thus f(x) is decreasing on \((-\infty, 0) \cup (0, 3)\). - \(f'(x) > 0\) for \(3 < x < \infty\), thus f(x) is increasing on \((3, \infty)\). Now, let's identify the extremas to analyze f'(x) at the critical points: - At \(x = 0\), \(f'(x)\) changes from negative to negative, so no relative extremum. - At \(x = 3\), \(f'(x)\) changes from negative to positive, so there is a relative minimum at \(x = 3\). To find the corresponding y-value, evaluate the original function: \(f(3) = (3)^4 - 4(3)^3 + 6 = 81 - 108 + 6 = -21\).
05

Final answer

a) The function f(x) is decreasing on \((-\infty, 0) \cup (0, 3)\) and increasing on \((3, \infty)\). b) There is a relative minimum at the point \((3, -21)\).

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Key Concepts

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

Critical Points
In calculus, the concept of critical points is fundamental for understanding the behavior of functions. Critical points are where the function's first derivative is either zero or undefined. For a given function, like our polynomial function f(x) = x^4 - 4x^3 + 6, we find the critical points by setting the first derivative f'(x) equal to zero.

In our exercise, by taking the derivative and factoring out common terms, we found that the critical points are at x = 0 and x = 3. Knowing these points helps us determine where the function's graph changes direction, which is essential for identifying intervals of increase and decrease, as well as any relative maxima or minima.
First Derivative Test
The First Derivative Test is a powerful tool to determine whether a function is increasing or decreasing around its critical points. The test involves analyzing the signs of the first derivative f'(x) before and after a critical point. If the derivative changes from positive to negative, the function has a relative maximum at that point. Conversely, if it changes from negative to positive, the function has a relative minimum.

For the function f(x), after finding that f'(x) is negative before and after x = 0 and changes from negative to positive at x = 3, we concluded that there is no relative extremum at x = 0, but there is a relative minimum at x = 3.

Using the First Derivative Test

Apply the test by choosing sample points in the intervals between critical points to determine the sign of f'(x). In this case, sample points such as x = -1, 1, and 4 show us the function's behavior in those respective intervals.
Relative Extremum
A relative extremum of a function is a point where the function reaches a local high or low value compared to its immediate surroundings. The First Derivative Test can help us identify such points by looking at the change in sign of the derivative. There are two types of relative extrema: relative maximum (the highest point in a region) and relative minimum (the lowest point in a region).

When applying this to our exercise, we found that at the critical point x = 3, the function f(x) has a relative minimum, because the first derivative changed from negative to positive. By plugging x = 3 back into the original function, we calculated the minimum value to be f(3) = -21. This comprehensive analysis grants us a better understanding of the function's graph, helping identify its rises and falls in different intervals, which is crucial for sketching the graph or finding optimal values in various applied mathematics problems.

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Most popular questions from this chapter

Approximating the \(k\) th Root of a Positive Number a. Apply Newton's method to the solution of the equation \(f(x)=x^{k}-A=0\) to show that an approximation of \(\sqrt[k]{A}\) can be found by using the iteration $$ x_{n+1}=\frac{1}{k}\left[(k-1) x_{n}+\frac{A}{x_{n}^{k-1}}\right] $$ b. Use this iteration to find \(\sqrt[10]{50}\) accurate to four decimal places.

Optimal Inventory Control The equation $$ A(q)=\frac{k m}{q}+c m+\frac{h q}{2} $$ gives the annual cost of ordering and storing (as yet unsold) merchandise. Here, \(q\) is the size of each order, \(k\) is the cost of placing each order, \(c\) is the unit cost of the product, \(m\) is the number of units of the product sold per year, and \(h\) is the annual cost for storing each unit. Determine the size of each order such that the annual cost \(A(q)\) is as small as possible.

Mass of a Moving Particle The mass \(m\) of a particle moving at a speed \(v\) is related to its rest mass \(m_{0}\) by the equation $$ m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ where \(c\), a constant, is the speed of light. Show that $$ \lim _{v \rightarrow c^{-}} \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\infty $$ thus proving that the line \(v=c\) is a vertical asymptote of the graph of \(m\) versus \(v .\) Make a sketch of the graph of \(m\) as a function of \(v\).

Worldwide PC Shipments The number of worldwide PC shipments (in millions of units) from 2005 through 2009 , according to data from the International Data Corporation, are given in the following table. \begin{tabular}{|l|c|c|c|c|c|} \hline Year & 2005 & 2006 & 2007 & 2008 & 2009 \\ \hline PCs & \(207.1\) & \(226.2\) & \(252.9\) & \(283.3\) & \(302.4\) \\ \hline \end{tabular} By using the logistic curve-fitting capability of a graphing calculator, it can be verified that a regression model for this data is given by $$ f(t)=\frac{544.65}{1+1.65 e^{-0.1846 r}} $$ where \(t\) is measured in years and \(t=0\) corresponds to 2005 . a. Plot the scatter diagram and the graph of the function \(f\) using the viewing window \([0,4] \times[200,300]\). b. How fast were the worldwide PC shipments increasing in \(2006 ?\) In \(2008 ?\)

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). \begin{array}{l} \text { The zero of } f(x)=x^{3}+2 x^{2}+x-6 \text { between } x=1 \text { and }\\\ x=2 . \text { Take } x_{0}=1.5 \text { . } \end{array}
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