Question

In Exercises \(19-30\) , find the extreme values of the function and where they occur. $$y=\frac{x+1}{x^{2}+2 x+2}$$

Short answer

The function \(y=\frac{x+1}{x^2+2x+2}\) has a global minimum at \(y=0\) as \(x -> \pm \inf\), but no local minimum or maximum, because the derivative of the function has no real roots.

Step-by-step solution

Find the derivative

Using the quotient rule, which states that the derivative of \( \frac{u}{v}\) is \(\frac{v \cdot u' - u \cdot v'}{v^2}\), where \(u\) is the numerator and \(v\) is the denominator and \(u'\) and \(v'\) are their respective derivatives. The derivative of the given function, \(y\), can be found as follows: Let \(u=x+1\) and \(v= x^2+2x+2\), then \(u'=1\) and \(v' = 2x+2\). The derivative then becomes: \( y'= \frac{(x^2+2x+2)*1 - (x+1)*(2x+2)}{(x^2+2x+2)^2} = \frac{2}{(x^2+2x+2)^2} \)

Solve for the critical points

The critical points are found by setting the derivative equal to zero and solving for \(x\), and also where the derivative is undefined. However, since our derivative \(y'=\frac{2}{{(x^2+2x+2)^2}}\) is defined for all \(x\), we only need to solve for when \(y'=0\) for critical points. But the equation \(y'=0\) has no solution because the numerator of \(y'\) is not zero. Therefore, there are no critical points.

Find the local extrema

Since the function has no critical points, it can have no local minimum or maximum points. To find the global minimum or maximum, one must look at the endpoints and any asymptotes. In this case, the function has a horizontal asymptote at \(y=0\), which suggests the function has a global minimum at \(y=0\). However, since the function is not restricted in the domain, it has no global maximum.

Find the points of extremum

After examining the asymptote, we find that as \(x -> \pm\inf\), \(y -> 0\). Therefore, the global minimum is \(y=0\), but it does not occur at specific \(x\) values, instead, it approaches this value as \(x\) approaches positive or negative infinity.