Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\frac{1}{1+x^{2}} ; I=(-\infty, \infty) $$

Step-by-Step Solution

Answer

The maximum value is 1 and the minimum value is 0 on \( (-\infty, \infty) \).

Step 1: Find the Derivative

To find the critical points of the function, we first need to find the derivative of the given function. The function is \( g(x) = \frac{1}{1 + x^2} \). Use the quotient rule, where if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = 1 \) and \( v(x) = 1 + x^2 \), so \( u'(x) = 0 \) and \( v'(x) = 2x \). Thus, the derivative is \( g'(x) = \frac{-2x}{(1 + x^2)^2} \).

Step 2: Set Derivative to Zero

Critical points occur where the derivative is zero or undefined. Since \( g'(x) = \frac{-2x}{(1 + x^2)^2} \) is never undefined (as the denominator is always positive), we just set \( g'(x) = 0 \) to find zeros.\[ \frac{-2x}{(1 + x^2)^2} = 0 \]The numerator must be zero, so \(-2x = 0\), giving \( x = 0 \).

Step 3: Evaluate at Critical Point

Evaluate \( g(x) \) at \( x = 0 \) to find the function value. \[ g(0) = \frac{1}{1 + 0^2} = 1 \]

Step 4: Consider End Behavior

Since the interval is infinitely large, consider the behavior of \( g(x) \) as \( x \to \pm \infty \). \[ \lim_{x \to \pm \infty} g(x) = \lim_{x \to \pm \infty} \frac{1}{1+x^2} = 0 \]The maximum value is at \( x = 0 \).

Step 5: Conclusion about Extrema

From the evaluations, the critical point at \( x = 0 \) yields a function value of 1, which is the maximum value. As \( x \to \pm \infty \), \( g(x) \to 0 \), which is the minimum value. Therefore, the maximum value of \( g(x) \) is 1, and the minimum is 0.

Key Concepts

Quotient Rule

The Quotient Rule is a technique used in calculus to find the derivative of a function that is the quotient of two other functions. If you have a function written as a fraction, like \( f(x) = \frac{u(x)}{v(x)} \), the Quotient Rule comes into play. It helps us differentiate this kind of function. The formula is:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]Here's what each part means:

• \( u(x) \) is the numerator function.
• \( v(x) \) is the denominator function.
• \( u'(x) \) is the derivative of the numerator.
• \( v'(x) \) is the derivative of the denominator.

For example, in the function \( g(x) = \frac{1}{1 + x^2} \), we see that:- The top (numerator) is \( 1 \) with derivative \( u'(x) = 0 \).- The bottom (denominator) is \( 1 + x^2 \) with derivative \( v'(x) = 2x \).Using these derivatives, we plug them into the Quotient Rule. We found that the derivative \( g'(x) = \frac{-2x}{(1 + x^2)^2} \). This result is fundamental as it helps determine where the function is increasing or decreasing.

Maximum and Minimum Values

Identifying the maximum and minimum values of a function involves several steps. These points are also known as extrema and occur where the derivative is zero or undefined, marking potential transition points between increasing and decreasing behavior.### Finding Critical PointsTo find these critical points, locate where the derivative equals zero or is undefined. In our function \( g(x) = \frac{1}{1 + x^2} \):

• The derivative \( g'(x) \) was \( \frac{-2x}{(1 + x^2)^2} \).
• This is never undefined because the denominator is always positive.
• The numerator \(-2x\) equals zero when \( x = 0 \).

Thus, \( x = 0 \) is a critical point.### Evaluating ExtremaNow, evaluate the function at \( x = 0 \):- Substitute \( x = 0 \) into \( g(x) \): \( g(0) = \frac{1}{1+0^2} = 1 \).This result indicates a maximum value since this is the highest point the function reaches. As for the minimum value, consider the function's behavior as \( x \to \pm \infty \), which leads us to discuss the end behavior.

End Behavior

The end behavior of a function examines what happens to the function's values as the input (\( x \)) heads to infinity or negative infinity. This is essential for functions defined across all real numbers or very large intervals, as it describes their behavior towards the extremes.For the function \( g(x) = \frac{1}{1 + x^2} \), consider:

• As \( x \to \infty \), the denominator \( 1 + x^2 \) becomes very large, making \( \frac{1}{1 + x^2} \) very small.
• Similarly, as \( x \to -\infty \), the behavior is the same because \( x^2 \) is always positive.
• Thus, both \( \lim_{x \to \pm \infty} g(x) = 0 \).

This analysis defines the minimum value of the function over its domain. The function approaches zero, but never actually reaches or goes below it. In summary, as x tends toward either very large positive or negative values, \( g(x) \) approaches its minimum possible value of 0.