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 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 \(u^\prime = 1\).
The derivative \(y^\prime=\frac{1}{2}(x + 1)^{-\frac{1}{2}}\times1=\frac{1}{2\sqrt{x+1}}\).

Step2: Analyze critical points

Set \(y^\prime = 0\), \(\frac{1}{2\sqrt{x + 1}}=0\). There is no solution for \(x\) since the numerator is non - zero for all real \(x\) in the domain \(2\leq x\leq10\).
Also, check the endpoints.
When \(x = 2\), \(y=\sqrt{2 + 1}=\sqrt{3}\).
When \(x = 10\), \(y=\sqrt{10 + 1}=\sqrt{11}\).
Since \(y^\prime=\frac{1}{2\sqrt{x+1}}>0\) for all \(x\in[2,10]\) (because the denominator \(2\sqrt{x + 1}>0\) when \(x\geq2\)).

Answer:

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