91Ó°ÊÓ

Critical points are essential when analyzing a function for extreme values. They occur where the derivative of a function is zero or undefined. In this exercise, to find the critical points of the function \( g(x) = \frac{x-2}{x^2-1} \), we take the first derivative using the quotient rule. Solving the resulting equation \( -x^2 -2x + 1 = 0 \), we obtain specific values of \( x \). By solving this quadratic equation, we find the potential critical points. However, not all of these values lie within our specified domain, \( 0 \leq x < 1 \). Thus, from our calculations, only the point \( x = \sqrt{2}-1 \) qualifies as a critical point within the domain.

The First Derivative Test helps us classify the critical points as local minima or maxima by examining the sign changes of the derivative around those points. If the derivative changes from positive to negative at a point, we have a local maximum. If it changes from negative to positive, it's a local minimum.
To perform the First Derivative Test for our function at \( x = \sqrt{2}-1 \), we examine values just below and above this point, such as \( x = 0.3 \) and \( x = 0.9 \). Calculating \( g'(0.3) \) and \( g'(0.9) \) allows us to check the sign change. When done correctly, if the derivative's sign changes as expected, this confirms the location and type of local extremum. For the given function, this approach reveals a local extremum at \( x = \sqrt{2}-1 \).

The Quotient Rule is a fundamental tool for differentiation used when dealing with functions represented as fractions. The formula for the Quotient Rule is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)

where \( u \) and \( v \) are functions of \( x \).
In our exercise, the function \( g(x) = \frac{x-2}{x^2-1} \) is differentiated using the Quotient Rule. Here, \( u = x - 2 \) and \( v = x^2 - 1 \). Applying the rule, we find that:

• \( g'(x) = \frac{(x^2-1)(1)-(x-2)(2x)}{(x^2-1)^2} \)
• Simplifying this expression, we get \( g'(x) = \frac{-x^2 - 2x + 1}{(x^2 - 1)^2} \)

This allows us to identify critical points by setting \( g'(x) = 0 \).

The Second Derivative Test provides another method to classify critical points as local minima or maxima. We take the second derivative of the function and evaluate it at the critical points we found.
If the second derivative, \( g''(x) \), is positive at a critical point, the function has a local minimum there. If it is negative, the point is a local maximum. If \( g''(x) = 0 \), other tests may be needed to determine the point's nature.
In our function \( g(x) = \frac{x-2}{x^2-1} \), computing the second derivative \( g''(x) \) and evaluating it at the critical point \( x = \sqrt{2} - 1 \) can offer a second perspective to confirm our earlier findings from the First Derivative Test. However, this exercise skips manual computation of \( g''(x) \), opting for the First Derivative Test for simplicity.