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The maximum value of y=√(x+2)-1 on the interval [-2 ,2]

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I love calculus! Let’s begin with finding the derivative of y.

y’ = (1/2)(x+2)^(-1/2)(1)
y’ = 1/(2 √(x+2))

Now to find the maximum or minimum, you need to check the end-points of the interval and when y’=0.

There is no instance where y’ will be equal to 0, so you just need to check the endpoints.

When x= -2, y= -1. When x= 2, y= 1.

There is an absolute maximum of 1 when x = 2

There is an absolute minimum of -1 when x = -2
User Alina
by
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4 votes

Answer:

1

Explanation:

The square root function is monotonically increasing, so will have a maximum at the right end of any interval. Here the maximum is at x=2. That value is ...

y = √(2+2) -1 = 2 -1 = 1

The maximum value on the interval is 1.

User NewShelbyWoo
by
7.1k points

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