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Critical points are essential in calculus as they help identify where a function may have relative extreme values like peaks or troughs. A critical point occurs specifically where the derivative of a function equals zero or becomes undefined. It's where the function's slope changes, indicating potential relative maxima or minima. For the function in our exercise, \[ g(x) = \frac{1}{x^2 + 1} \], we first find the derivative using the quotient rule.

• Set the derivative, \( g'(x) = -\frac{2x}{(x^2 + 1)^2} \), equal to zero to find critical points.
• Solve the equation to get \( x = 0 \) as the critical point.
• Check if the derivative is undefined anywhere; it's not here because \((x^2 + 1)^2 eq 0\) for all real \( x \).

In this function, the analysis shows that the only critical point is at \( x = 0 \). This result sets the stage for further study to determine what happens around this point.

When dealing with complex functions involving division, the quotient rule is a powerful tool to compute derivatives. It's especially useful when a function is a fraction of two different functions. For a function given as \( \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the quotient rule states:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. \]In our exercise, \( g(x) = \frac{1}{x^2 + 1} \), the numerator is 1 (a constant function), and the denominator is \( x^2 + 1 \).

• The derivative of the numerator, \( 1 \), is 0.
• The derivative of the denominator, \( x^2 + 1 \), is \( 2x \).

Applying the quotient rule then gives us:\[ g'(x) = \frac{0(x^2 + 1) - 1(2x)}{(x^2 + 1)^2} = -\frac{2x}{(x^2 + 1)^2}. \]Understanding this rule simplifies determining the slope of \( g(x) \) at various points and helps pinpoint potential critical points.

Determining relative extreme values is crucial since these points reflect peaks and valleys of a function—that is, maxima and minima. Relative extreme values are observed at critical points where there is a change in slope. The First Derivative Test helps determine these changes by analyzing the sign of the function's derivative at points around the critical points.For the function \( g(x) = \frac{1}{x^2 + 1} \), with a derivative:\[ g'(x) = -\frac{2x}{(x^2 + 1)^2}, \]we analyze the changes in \( g'(x) \) around the critical point \( x = 0 \).

• For \( x < 0 \), \( g'(x) \) is positive indicating an upward slope.
• For \( x > 0 \), \( g'(x) \) is negative indicating a downward slope.

Since \( g'(x) \) transitions from positive to negative, \( x = 0 \) is identified as a relative maximum. Evaluating \( g(x) \) at this point, we find:\[ g(0) = \frac{1}{0^2 + 1} = 1. \]Thus, there is a relative maximum at \( x = 0 \) with a function value of 1. Such analysis is vital for understanding the behavior of functions in various applications.