/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Find those values of \(x\) for w... [FREE SOLUTION] | 91Ó°ÊÓ

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Find those values of \(x\) for which the given functions are increasing and those values of \(x\) for which they are decreasing. $$y=2+27 x-x^{3}$$

Short Answer

Expert verified
The function is increasing on \((-3, 3)\) and decreasing on \((-\infty, -3) \cup (3, \infty)\).

Step by step solution

01

Differentiate the function

To find where the function is increasing or decreasing, we first need to find its derivative. The given function is \(y = 2 + 27x - x^3\). Differentiate the function with respect to \(x\):\[\frac{dy}{dx} = 27 - 3x^2.\]
02

Determine the critical points

Set the derivative equal to zero to find the critical points, where the tangent is horizontal (neither increasing nor decreasing):\[27 - 3x^2 = 0.\]Solve for \(x\) by rearranging and dividing:\[3x^2 = 27 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3.\]
03

Test intervals around critical points

Test the intervals around the critical points \(-3\) and \(3\) using test values. Pick \(x = -4\), \(x = 0\), and \(x = 4\), and substitute these into the derivative \(27 - 3x^2\):- For \(x = -4\), \(\frac{dy}{dx} = 27 - 3(-4)^2 = 27 - 48 = -21\) (decreasing)- For \(x = 0\), \(\frac{dy}{dx} = 27 - 3(0)^2 = 27 \) (increasing)- For \(x = 4\), \(\frac{dy}{dx} = 27 - 3(4)^2 = 27 - 48 = -21\) (decreasing)
04

Summarize intervals of increase and decrease

Based on the sign changes of the derivative in each interval, we conclude:- The function is **decreasing** on \((-\infty, -3)\) and \((3, \infty)\) because the derivative is negative.- The function is **increasing** on \((-3, 3)\) because the derivative is positive.

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Key Concepts

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

Derivative
The derivative of a function is like a speedometer for your car. Instead of measuring miles per hour, it measures how fast the function is changing.
For the function given as - \( y = 2 + 27x - x^3 \),we take the derivative with respect to \( x \) to find out how the function's output changes as \( x \) changes. The derivative, - \( \frac{dy}{dx} = 27 - 3x^2 \),tells us the rate of change of the function at any point \( x \).
  • A positive derivative means the function is increasing at that point (just like moving forward in a car).
  • A negative derivative means the function is decreasing (like moving backward).
Critical points
Critical points are special places where the function's journey changes course. These are the points where the derivative equals zero - \( 27 - 3x^2 = 0 \).To solve this equation, we rearrange it to find values of \( x \):- \( x^2 = 9 \),leading to the solutions- \( x = \pm 3 \).These values mark the spots where the function could switch from increasing to decreasing, or vice versa. You can think of critical points as potential turning points or flat spots on our journey.
Intervals of increase and decrease
Intervals of increase and decrease tell us where the function is climbing upward (increasing) or sliding downward (decreasing), which we find by testing the intervals divided by our critical points:- For \( x < -3 \), the derivative - \( \frac{dy}{dx} < 0 \), meaning the function is decreasing.- For \( -3 < x < 3 \), the derivative - \( \frac{dy}{dx} > 0 \), so the function is increasing.- For \( x > 3 \), the derivative - \( \frac{dy}{dx} < 0 \), which suggests the function is decreasing.Visualizing these intervals is like seeing when a roller coaster goes up or down, with the critical points as the places where the direction changes.
Polynomial function
A polynomial function, such as our given - \( y = 2 + 27x - x^3 \),is built from constants and variables raised to whole number powers. They can have curves and bends, but no breaks or sharp corners. In mathematics, polynomial functions are vital due to their simplicity and the ease with which we can compute their derivatives. The degree of the polynomial tells you the highest power of \( x \); here it's a third-degree polynomial because of the \( x^3 \) term.
The behavior of the polynomial function across its domain can be effectively analyzed using derivatives, allowing us to determine where it rises or falls. These functions often appear in various real-world and theoretical contexts, making them an important study area in calculus.

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Most popular questions from this chapter

In Exercises \(21-32,\) sketch the graphs of the given functions by determining the appropriate information and points from the first and second derivatives. Use a calculator to check the graph. In Exercises \(27-32,\) use the function maximum-minimum feature to check the local maximum and minimum points. $$y=x(x-4)^{3}$$

Solve the given maximum and minimum problems. Computer simulation shows that the drag \(F\) (in \(\mathrm{N}\) ) on a certain airplane is \(F=0.00500 v^{2}+3.00 \times 10^{8} / v^{2},\) where \(v\) is the velocity (in \(\mathrm{km} / \mathrm{h}\) ) of the plane. For what velocity is the drag the least?

In Exercises \(43-54,\) sketch the indicated curves by the methods of this section. You may check the graphs by using a calculator. An electric circuit is designed such that the resistance \(R\) (in \(\Omega\) ) is a function of the current \(i\) (in \(\mathrm{mA}\) ) according to \(R=75-18 i^{2}+8 i^{3}-i^{4} .\) Sketch the graph if \(R \geq 0\) and \(i\) can be positive or negative.

Solve the given problems by finding the appropriate differential. The velocity of an object rolling down a certain inclined plane is given by \(v=\sqrt{100+16 h},\) where \(h\) is the distance (in \(\mathrm{ft}\) ) traveled along the plane by the object. What is the increase in velocity (in \(\mathrm{ft} / \mathrm{s}\) ) of an object in moving from \(20.0 \mathrm{ft}\) to \(20.5 \mathrm{ft}\) along the plane? What is the relative change in the velocity?

Computer simulation shows that the drag \(F\) (in \(\mathrm{N}\) ) on a certain airplane is \(F=0.00500 v^{2}+3.00 \times 10^{8} / v^{2},\) where \(v\) is the velocity (in \(\mathrm{km} / \mathrm{h}\) ) of the plane. For what velocity is the drag the least?

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