+0  
 
0
45
3
avatar

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 $f(3)$?

Guest Mar 14, 2018
Sort: 

3+0 Answers

 #1
avatar+6945 
+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

hectictar  Mar 15, 2018
 #3
avatar+85821 
+1

Nice one, hectictar   !!!

 

 

cool cool cool

CPhill  Mar 15, 2018
 #2
avatar+85821 
+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 

CPhill  Mar 15, 2018

40 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details