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Chapter 4: Problem 16

Sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem 1. $$y=\frac{6}{x^{2}+2}, \quad-1 < x < 1$$

Short Answer

Expert verified
The function reaches an absolute maximum of 3 at \(x = 0\) and approaches a minimum near boundaries \(y \to 2\).

Step by step solution

01

Analyze Domain and Range

The function given is \( y = \frac{6}{x^2 + 2} \). The domain is \(-1 < x < 1\), meaning \( x \) can take any value between \(-1\) and \(1\) but not including the endpoints. Since \(x^2 + 2\) is always positive for any real \( x \), the function is defined for the entire interval and has a range \(0 < y \leq 3\) as the minimum value of \(x^2 + 2\) is \(2\), giving a maximum for \(y\).
02

Determine Critical Points

To find critical points where absolute extreme values may occur, we'll differentiate \(f(x) = \frac{6}{x^2 + 2}\). \[ f'(x) = -\frac{12x}{(x^2 + 2)^2} \]Set \( f'(x) = 0 \) to find critical points, which yields \( 12x = 0 \), giving \( x = 0 \). This point is within the domain and needs evaluation for extremum.
03

Evaluate y at Critical Points

Evaluate the function at the critical point found:\[ f(0) = \frac{6}{0^2 + 2} = 3 \] This is the maximum value within the domain as \(f(x)\) decreases symmetrically on either side of 0.
04

Visualize the Behavior Near Boundaries

As \( x \) approaches the boundary points, \(-1\) and \(1\), of the domain, \(x^2\) approaches 1. Evaluate \[ \lim_{x \to 1^-} f(x) = \frac{6}{3} = 2 \\lim_{x \to -1^+} f(x) = \frac{6}{3} = 2 \]Thus, \( y \) approaches 2 at both endpoints, but never actually reaches them, staying consistent with \(y > 2\).
05

Consistency with Theorem 1

Theorem 1 states a continuous function on a closed interval attains absolute maxima and minima. Although our interval is open, we found the local maximum at \(x = 0\), consistent with the domain limits approaching the asymptotic \(y \approx 2\) near the boundaries.

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Key Concepts

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

Extreme Values
Extreme values are the highest and lowest values that a function can take within a given interval. These values are often referred to as maxima and minima. In the case of the function given in the exercise, we explore the absolute extreme values within the domain \(-1 < x < 1\). While the function is defined throughout this interval, especially because \(x^2 + 2\) is positive for any real \(x\), it is vital to examine how the function behaves to identify extreme values.
  • Absolute maximum is observed at \(x = 0\), where \(f(x) = 3\).
  • The function does not attain an absolute minimum within the open interval.
  • Near the boundary points, \(x = -1\) and \(x = 1\), the function approaches \(y = 2\), but never reaches it.
Therefore, the maximum value of the function in its domain is 3, but an absolute minimum value isn't achieved in the open interval as the function value approaches but doesn't equal 2.
Function Domain
The domain of a function includes all the possible input values (x-values) for which the function is defined. For the function \(y=\frac{6}{x^2 + 2}\), the domain is given as \(-1 < x < 1\). This interval is open, meaning \(x\) can take any value strictly between \(-1\) and \(1\), without including the endpoints.
  • The expression \(x^2 + 2\) guarantees positivity, hence the function remains well-defined for all \(x\) in the given domain.
  • The function experiences no discontinuities or undefined points within this range, as the denominator never approaches zero.
  • Within this domain, the function explores a range of output values \(0 < y \leq 3\).
Understanding the function's domain is crucial for determining where the function is applicable and predicting the behavior related to extreme values.
Critical Points
Critical points are locations on a graph where the function's derivative is either zero or undefined. These points help in finding potential maxima or minima. To locate critical points of \(y = \frac{6}{x^2 + 2}\), we first find the derivative and solve for zero.
  • The derivative is \(-\frac{12x}{(x^2 + 2)^2}\).
  • Setting the derivative equal to zero, \(12x = 0\), we find \(x = 0\) as a critical point.
  • This value lies within the domain \(-1 < x < 1\) and therefore is evaluated for being an extremum point.
By identifying critical points, we can apply further analysis to determine where extreme values might exist within the function's domain.
Derivative Analysis
Derivative analysis involves evaluating the derivative of a function to understand its behavior and changes in slope. For the function \(y = \frac{6}{x^2 + 2}\), deriving the function helps us identify slopes and m trends.
  • The derivative, \(f'(x) = -\frac{12x}{(x^2 + 2)^2}\), indicates the rate of change of \(y\) with respect to \(x\).
  • The sign of the derivative provides insights: \(f'(x) > 0\) implies an increasing function, while \(f'(x) < 0\) indicates a decreasing function.
  • At the critical point \(x = 0\), the derivative is zero, suggestive of a possible local extremum.
  • Near \(x = 0\), the function exhibits symmetry, decreasing as \(x\) moves away from zero within the domain.
By studying derivatives, we can better predict the graph's shape and locate points where the function achieves its extreme values.

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Most popular questions from this chapter

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int x^{-5 / 4} d x$$

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