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Question

Use the equation y=1+xy=1+\sqrt{x} to answer the following questions.

(a) For what values of x is y = 4?

(b) For what values of x is y = 0?

(c) For what values of x is y ≥ 6?

(d) Does y have a minimum value? A maximum value? If so, find them.

Solution

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(a)\textbf{(a)}

When y=4y=4,

4=1+x4=1+\sqrt{x}

3=x3=\sqrt{x}

Square both sides:

9=x9=x

x=9\color{#c34632}x=9

(b)\textbf{(b)}

When y=0y=0,

0=1+x0=1+\sqrt{x}

1=x-1=\sqrt{x}

The right side cannot be negative so there are no xx-values that make y=0y=0.

(c)\textbf{(c)}

Solve for the inequality:

1+x61+\sqrt{x}\geq 6

x5\sqrt{x}\geq 5

Square both sides:

x25\color{#c34632}x\geq 25

(d)\textbf{(d)}

Yes. y=1+xy=1+\sqrt{x} has a domain of x0x\geq 0 and is increasing so its minimum occurs at x=0x=0:

y=1+0y=1+\sqrt{0}

y=1+0y=1+0

y=1y=\color{#c34632}1

Since it is increasing on its domain, then it has no maximum value.

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