Domain calculations help needed!

Kendall

New member
Joined
May 3, 2013
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10
Hello,
I apologise if I posted in the wrong section...

I have a problem to find the correct domain, here is what I have done so far:
F(x)=sqrt7x-4 and g(x)=3x^2-1 Find H(x)

h(x) = (f o f)(x)
h(x)=sqrt(7*sqrt (7*x+(-4))+(-4))This is also correct I checked with Professor

Domain
I am not sure how to do it because it is double root. so here is what I did, but it is wrong
[0.571428571, Inf)


Please, help me to get to the correct answer!
Thank you I apologise if I posted in the wrong section
 
First we know that 7 f - 4 must be non-negative since we are taking the square root of that. So f is no smaller than 4/7 or
(7x4)47\displaystyle \sqrt(7 x - 4) \ge \frac{4}{7}
=> 7x41649=(47)2\displaystyle 7 x - 4 \ge \frac{16}{49}=(\frac{4}{7})^2
=> 7x1649+4\displaystyle 7 x \ge \frac{16}{49}+4
=> 7x21249\displaystyle 7 x \ge \frac{212}{49}
=> x212343\displaystyle x \ge \frac{212}{343}

Next 7 x - 4 must be non-negative since we are taking the square root of that. So x is no smaller than 4/7 or
x47=196343\displaystyle x \ge \frac{4}{7} = \frac{196}{343}

Since x must satisfy both of those equations,
x212343\displaystyle x \ge \frac{212}{343} ~ 0.6181
and the domain is [212343\displaystyle \frac{212}{343}, ∞)

Edited to correct arithmetic. BTW: I would really check this, I don't trust myself to do simple arithmetic today for some reason.
 
Last edited:
find the correct domain....
F(x)=sqrt7x-4 and g(x)=3x^2-1 Find H(x)
How is "H(x)" defined? Note: The function "H(x)" is different from the function "h(x)". Also, is "F(x)" equal to any of the following?

. . . . .\(\displaystyle \mbox{a) }\, F(x)\, =\, \sqrt{7\, }x\, -\, 4\)

. . . . .\(\displaystyle \mbox{b) }\, F(x)\, =\, \sqrt{7x\, }\, -\, 4\)

. . . . .\(\displaystyle \mbox{c) }\, F(x)\, =\, \sqrt{7x\, -\, 4\, }\)

Or something else?

h(x) = (f o f)(x)
h(x)=sqrt(7*sqrt (7*x+(-4))+(-4))This is also correct I checked with Professor
How do "h(x)" and "f(x)" relate to the original functions, "H(x)" and "F(x)"?

Domain
I am not sure how to do it because it is double root. so here is what I did, but it is wrong
[0.571428571, Inf)
How did you arrive at this value? Please reply showing your steps. Thank you! ;)
 
First we know that 7 f - 4 must be non-negative since we are taking the square root of that. So f is no smaller than 4/7 or
(7x4)47\displaystyle \sqrt(7 x - 4) \ge \frac{4}{7}
=> 7x41649=(47)2\displaystyle 7 x - 4 \ge \frac{16}{49}=(\frac{4}{7})^2
=> 7x1649+4\displaystyle 7 x \ge \frac{16}{49}+4
=> 7x21249\displaystyle 7 x \ge \frac{212}{49}
=> x212343\displaystyle x \ge \frac{212}{343}

Next 7 x - 4 must be non-negative since we are taking the square root of that. So x is no smaller than 4/7 or
x47=196343\displaystyle x \ge \frac{4}{7} = \frac{196}{343}

Since x must satisfy both of those equations,
x212343\displaystyle x \ge \frac{212}{343} ~ 0.6181
and the domain is [212343\displaystyle \frac{212}{343}, ∞)

Edited to correct arithmetic. BTW: I would really check this, I don't trust myself to do simple arithmetic today for some reason.
Hi, thank you so much! Just check with Prof, your calculations are correct!!!
 
How is "H(x)" defined? Note: The function "H(x)" is different from the function "h(x)". Also, is "F(x)" equal to any of the following?

. . . . .\(\displaystyle \mbox{a) }\, F(x)\, =\, \sqrt{7\, }x\, -\, 4\)

. . . . .\(\displaystyle \mbox{b) }\, F(x)\, =\, \sqrt{7x\, }\, -\, 4\)

. . . . .\(\displaystyle \mbox{c) }\, F(x)\, =\, \sqrt{7x\, -\, 4\, }\)

Or something else?

How do "h(x)" and "f(x)" relate to the original functions, "H(x)" and "F(x)"?

How did you arrive at this value? Please reply showing your steps. Thank you! ;)
Thanks, I got where I was wrong, here is new calculation
(f o f)(x) = f(f(x)) = [FONT=MathJax_Main-Web]f[/FONT][FONT=MathJax_Main-Web]([/FONT][FONT=MathJax_Main-Web]7x[/FONT][FONT=MathJax_Main-Web]−[/FONT][FONT=MathJax_Main-Web]4[/FONT]
newreply.php
[FONT=MathJax_Main-Web]−
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[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
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[/FONT]
newreply.php
[FONT=MathJax_Main-Web]√[/FONT]
newreply.php
[FONT=MathJax_Main-Web])[/FONT]
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requires [FONT=MathJax_Main-Web]7x[/FONT][FONT=MathJax_Main-Web]−[/FONT][FONT=MathJax_Main-Web]4[/FONT]
newreply.php
[FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT][FONT=MathJax_Main-Web]−
newreply.php
[/FONT]
newreply.php
[FONT=MathJax_Main-Web]√[/FONT]
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[FONT=MathJax_Main-Web]≥[/FONT][FONT=MathJax_Main-Web]4[/FONT][FONT=MathJax_Main-Web]/[/FONT][FONT=MathJax_Main-Web]7[/FONT]
newreply.php
, i.e. [FONT=MathJax_Main-Web]7[/FONT][FONT=MathJax_Main-Web]x[/FONT][FONT=MathJax_Main-Web]≥[/FONT][FONT=MathJax_Main-Web]212[/FONT][FONT=MathJax_Main-Web]/[/FONT][FONT=MathJax_Main-Web]49[/FONT]
newreply.php
,
so its domain is [212/343, ∞) ⋂ [4/7, ∞),
which is [212/343, ∞).
 
How is "H(x)" defined? Note: The function "H(x)" is different from the function "h(x)". Also, is "F(x)" equal to any of the following?

. . . . .\(\displaystyle \mbox{a) }\, F(x)\, =\, \sqrt{7\, }x\, -\, 4\)

. . . . .\(\displaystyle \mbox{b) }\, F(x)\, =\, \sqrt{7x\, }\, -\, 4\)

. . . . .\(\displaystyle \mbox{c) }\, F(x)\, =\, \sqrt{7x\, -\, 4\, }\)

Or something else?


How do "h(x)" and "f(x)" relate to the original functions, "H(x)" and "F(x)"?


How did you arrive at this value? Please reply showing your steps. Thank you! ;)
Thank you for your help, I got the answers:
sqrt 7x-4> or =4/7 then 7x>= 212/49 so domain is [212/343, infinity)
 
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