Question

find the local maxima and minima for the function. find the intervals on which it is increasing and the intervals on which it is decreasing.

( y = sqrt { x + 1 } , 2 leq x leq 10 )

find the local maxima and minima of the function. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
(type an ordered pair. type an exact answer for each coordinate, using radicals as needed. use a comma to separate answers as needed.)

a. the local minimum/minima is/are and the local maximum/maxima is/are

b. the local minimum/minima is/are ( ( 2 , sqrt { 3 } ) ) and there are no local maxima.

c. the local maximum/maxima is/are and there are no local minima.

d. there are no local minima or maxima.

Explanation:

Step1: Find the derivative

The function is \(y = \sqrt{x + 1}=(x + 1)^{\frac{1}{2}}\). Using the power rule \((u^n)^\prime=nu^{n - 1}u^\prime\), where \(u=x + 1\), \(n=\frac{1}{2}\), and \(u^\prime=1\). So \(y^\prime=\frac{1}{2}(x + 1)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x+1}}\).

Step2: Analyze critical points

Critical points occur where \(y^\prime = 0\) or \(y^\prime\) is undefined. Since \(y^\prime=\frac{1}{2\sqrt{x + 1}}\), \(y^\prime\) is never \(0\) (the numerator is \(1
eq0\)). \(y^\prime\) is undefined when \(x+1 = 0\) (i.e., \(x=-1\)), but \(x=-1
otin[2,10]\).

Step3: Analyze the function on the interval \([2,10]\)

Since \(y^\prime=\frac{1}{2\sqrt{x + 1}}>0\) for all \(x\in[2,10]\) (because the numerator \(1>0\) and the denominator \(2\sqrt{x + 1}>0\) for \(x\in[2,10]\)), the function \(y=\sqrt{x + 1}\) is increasing on the interval \([2,10]\).

Answer:

B. The local minimum/minima is/are \((2,\sqrt{3})\) and there are no local maxima.