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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
 #1
avatar+7612 
+2

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
avatar+100571 
+1

Nice one, hectictar   !!!

 

 

cool cool cool

CPhill  Mar 15, 2018
 #2
avatar+100571 
+2

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

 

 

cool cool cool 

 Mar 15, 2018

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