**How do I answer this Calculus 1 fct problem?**

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*y=1+sqrt x*

when the values of x is y=4?

for what value of x is y= 0?

for what value of x is y greater than or equal to 6?

y=1+√x

x≥0 ==> Domain=[0,∞);

x=0 ==> y=1=ymin ==> Range=[1,∞)

1) y=4 ==> 4=1+√x ==> √x=3 ==> x=9

2) y=0 is impossible (out of Range)

3) y≥6 ==> 1+√x≥6 ==> √x≥5 ==> x≥5² ==> x∈[25,∞)

**Answer’s Too Low**