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# Use the equation $y=1+\sqrt{x}$ 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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$\textbf{(a)}$

When $y=4$,

$4=1+\sqrt{x}$

$3=\sqrt{x}$

Square both sides:

$9=x$

$\color{#c34632}x=9$

$\textbf{(b)}$

When $y=0$,

$0=1+\sqrt{x}$

$-1=\sqrt{x}$

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

$\textbf{(c)}$

Solve for the inequality:

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

$\sqrt{x}\geq 5$

Square both sides:

$\color{#c34632}x\geq 25$

$\textbf{(d)}$

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

$y=1+\sqrt{0}$

$y=1+0$

$y=\color{#c34632}1$

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

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