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Chapter 4: Problem 3

Find the critical points for each function. Use the first derivative test to determine whether the critical point is a local maximum, local minimum, or neither. a. \(y=x^{4}-8 x^{2}\) b. \(f(x)=\frac{2 x}{x^{2}+9} \quad\) c. \(y=x^{3}+3 x^{2}+1\)

Short Answer

Expert verified
a. Local min at \( x = -2, 2 \), local max at \( x = 0 \). b. Local max at \( x = -3, 3 \). c. Local max at \( x = -2 \), local min at \( x = 0 \).

Step by step solution

01

Differentiate the given function (a)

The function given is \( y = x^4 - 8x^2 \). First, we find the derivative using the power rule. The derivative is \( y' = 4x^3 - 16x \).
02

Find critical points for function (a)

Set the derivative equal to zero to find critical points: \( 4x^3 - 16x = 0 \). Factor out the greatest common factor, \( 4x(x^2 - 4) = 0 \). This equation can be further factored to \( 4x(x-2)(x+2) = 0 \), yielding critical points at \( x = 0, -2, 2 \).
03

Determine the nature of critical points for function (a) using the First Derivative Test

Examine intervals around the critical points: - For \( x < -2 \), choose \( x = -3 \): \( y' = 4(-3)^3 - 16(-3) = -108 + 48 = -60 \), function is decreasing. - For \( -2 < x < 0 \), choose \( x = -1 \): \( y' = 4(-1)^3 - 16(-1) = -4 + 16 = 12 \), function is increasing.- For \( 0 < x < 2 \), choose \( x = 1 \): \( y' = 4(1)^3 - 16(1) = 4 - 16 = -12 \), function is decreasing.- For \( x > 2 \), choose \( x = 3 \): \( y' = 4(3)^3 - 16(3) = 108 - 48 = 60 \), function is increasing.Thus, \( x = -2 \) is a local minimum, \( x = 0 \) is a local maximum, and \( x = 2 \) is a local minimum.
04

Differentiate the given function (b)

The function is \( f(x) = \frac{2x}{x^2 + 9} \). By applying the quotient rule, the derivative is \( f'(x) = \frac{(x^2 + 9)2 - 2x(2x)}{(x^2 + 9)^2} = \frac{18 - 2x^2}{(x^2 + 9)^2} \).
05

Find critical points for function (b)

Set the numerator of the derivative equal to zero: \( 18 - 2x^2 = 0 \). Solving for \( x \), we have \( 2x^2 = 18 \) leading to \( x^2 = 9 \). Hence, the critical points are \( x = 3 \) and \( x = -3 \).
06

Determine the nature of critical points for function (b) using the First Derivative Test

Examine signs around critical points:- For \( x < -3 \), choose \( x = -4 \): \( f'(-4) = \frac{18 - (-4)^2 \cdot 2}{((-4)^2 + 9)^2} = \frac{18 - 32}{289} \), function is decreasing.- For \( -3 < x < 3 \), choose \( x = 0 \): \( f'(0) = \frac{18}{81} > 0 \), function is increasing.- For \( x > 3 \), choose \( x = 4 \): \( f'(4) = \frac{18 - 32}{289} \), function is decreasing. So, both \( x = -3 \) and \( x = 3 \) are local maxima.
07

Differentiate the given function (c)

The function is \( y = x^3 + 3x^2 + 1 \). Find the derivative: \( y' = 3x^2 + 6x \).
08

Find critical points for function (c)

Set the derivative equal to zero: \( 3x^2 + 6x = 0 \). Factor out the greatest common factor: \( 3x(x + 2) = 0 \). The critical points are \( x = 0 \) and \( x = -2 \).
09

Determine the nature of critical points for function (c) using the First Derivative Test

Examine intervals around the critical points:- For \( x < -2 \), choose \( x = -3 \): \( y'(-3) = 3(-3)^2 + 6(-3) = 27 - 18 = 9 \), function is increasing.- For \( -2 < x < 0 \), choose \( x = -1 \): \( y'(-1) = 3(-1)^2 + 6(-1) = 3 - 6 = -3 \), function is decreasing.- For \( x > 0 \), choose \( x = 1 \): \( y'(1) = 3(1)^2 + 6(1) = 3 + 6 = 9 \), function is increasing. Thus, \( x = -2 \) is a local maximum and \( x = 0 \) is a local minimum.

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Key Concepts

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

First Derivative Test
The First Derivative Test is a valuable tool in calculus for determining the nature of critical points of a function. These critical points occur where the derivative of the function is zero or undefined. Once you find these points, the First Derivative Test helps to classify them as a local maximum, local minimum, or neither.

Here's how it works:
  • Calculate the derivative of the function.
  • Find critical points by setting the derivative equal to zero.
  • Examine the sign of the derivative before and after each critical point.
  • If the derivative changes from positive to negative, the point is a local maximum.
  • If the derivative changes from negative to positive, the point is a local minimum.
  • If there is no change in sign, the point is a saddle point or inflection point, and not a local extremum.
This test provides an effective way to understand the behavior of functions and visualize their graphs.
Power Rule
The Power Rule is a basic yet essential technique in calculus for finding derivatives. It applies when differentiating functions in the form of polynomials. The rule states that the derivative of a term like \(x^n\) is \(nx^{n-1}\), where \(n\) is any real number.

Here鈥檚 a simple example:
  • If you have \(x^4\), applying the Power Rule gives you \(4x^3\) as the derivative.
  • Similarly, for \(-8x^2\), the derivative becomes \(-16x\).
This rule helps streamline finding derivatives for polynomial functions, allowing for quick computation and analysis without needing complex methods.
Quotient Rule
The Quotient Rule is used to find the derivative of functions that are fractions of two differentiable functions. It is a bit more involved than the Power Rule and is vital when dealing a situation where one function is divided by another.

The formula for the Quotient Rule is:
  • If you have two functions \(u\) and \(v\), and you want the derivative of \(\frac{u}{v}\), it's given by \(\left(\frac{u'}{v} - \frac{u \cdot v'}{v^2}\right)\).
For example, in the function \(f(x) = \frac{2x}{x^2 + 9}\):
  • The derivative is computed as \(\frac{(x^2 + 9) \cdot 2 - 2x(2x)}{(x^2 + 9)^2}\),
  • resulting in \(\frac{18 - 2x^2}{(x^2 + 9)^2}\).
This assists in analyzing functions that involve divisions, providing a clear approach to understanding their rates of change.
Local Maximum and Minimum
Local maxima and minima are key concepts in analyzing the behavior of functions. They indicate where the function reaches a high or low point in a particular section of the domain.

Here are some characteristics:
  • A local maximum is a point where the function value is higher than its immediate surroundings. A hilltop in a landscape.
  • A local minimum is where the function value is lower than its immediate surroundings. A valley in a landscape.
  • Use the First Derivative Test to confirm if a critical point is a local extremum.
  • The behavior before and after these points (whether increasing or decreasing) aids in identifying these peaks and valleys.
Understanding local maxima and minima is crucial for graphing functions and identifying peak efficiency or performance when analyzing real-world scenarios.

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Most popular questions from this chapter

Use a calculator to graph each of the following functions. Inspect the graph to estimate where the function is increasing and where it is decreasing. Verify your estimates with algebraic solutions. a. \(f(x)=x^{3}+3 x^{2}+1\) b. \(f(x)=x^{5}-5 x^{4}+100\) c. \(f(x)=x+\frac{1}{x}\) d. \(f(x)=\frac{x-1}{x^{2}+3}\) e. \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}\) f. \(f(x)=x^{4}+x^{2}-1\)

Given the following results of the analysis of a function, sketch a possible graph for the function: a. \(f(0)=0,\) the horizontal asymptote is \(y=2,\) the vertical asymptote is \(x=3,\) and \(f^{\prime}(x)<0\) and \(f^{\prime \prime}(x)<0\) for \(x<3 ; f^{\prime}(x)<0\) and \(f^{\prime \prime}(x)>0\) for \(x>3\) b. \(f(0)=6, f(-2)=0\) the horizontal asymptote is \(y=7,\) the vertical asymptote is \(x=-4,\) and \(f^{\prime}(x)>0\) and \(f^{\prime \prime}(x)>0\) for \(x<-4\) \(f^{\prime}(x)>0\) and \(f^{\prime \prime}(x)<0\) for \(x>-4\)

Determine the points at which \(f^{\prime}(x)=0\) for each of the following functions: a. \(f(x)=x^{3}+6 x^{2}+1\) b. \(f(x)=\sqrt{x^{2}+4}\) c. \(f(x)=(2 x-1)^{2}\left(x^{2}-9\right)\) d. \(f(x)=\frac{5 x}{x^{2}+1}\)

Find constants \(a\) and \(b\) such that the graph of the function defined by \(f(x)=\frac{a x+5}{3-b x}\) will have a vertical asymptote at \(x=5\) and a horizontal asymptote at \(y=-3\).

Find the value of the constant \(b\) such that the function \(f(x)=\sqrt{x+1}+\frac{b}{x}\) has a point of inflection at \(x=3\)
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