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# Need help

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1.  Compute the domain of the real-valued function $$f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}.$$

2. If $f(x)=ax^4-bx^2+x+5$ and $f(-3)=2$, then what is the value of $f(3)$?

Mar 14, 2018

### 3+0 Answers

#1
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1.   $$f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}$$

$$\sqrt{x}$$   is part of the function so the domain has the restriction:

x ≥ 0

$$\sqrt{5-\sqrt{x}}$$   is part of the function so the domain has the restriction:

$$5-\sqrt{x}\geq0\\ 5\geq\sqrt{x}\\ 25\geq x\\ \mathbf{x\leq 25}$$

$$\sqrt{3-\sqrt{5-\sqrt x}}$$   is part of the function so the domain has the restriction:

$$3-\sqrt{5-\sqrt x}\geq0\\ 3\geq\sqrt{5-\sqrt x}\\ 9\geq5-\sqrt x\\ 9+\sqrt x\geq5$$

$$\sqrt x\geq-4$$          This is true for all real numbers.

The domain restrictions are:  x ≥ 0  and  x ≤ 25

So the domain is just

0 ≤ x ≤ 25

Mar 15, 2018
#3
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Nice one, hectictar   !!!   CPhill  Mar 15, 2018
#2
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2. If f(x)=ax^4-bx^2+x+5 and f(-3)=2, then what is the value of f(3)?

If  f(-3)  = 2, we have

2  = a(-3)^4 - b(-3)^2 - 3 + 5

2  = 81a - 9b + 2

0 = 81a - 9b

9b  = 81a

b = 9a

f(3)  =  a(3)^4 - (9a)(3)^2 + 3 + 5

f(3)  = 81a - 81a  + 3 + 5

f(3)  = 8   Mar 15, 2018