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Range of a Function

  • Thread starter Codester09
  • Start date
  • #1
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Homework Statement



Find the range of h

Homework Equations



h(x) = sqrt(25 + (x - 3)2)

The Attempt at a Solution



I factored out the (x-3)2 and simplified to get

sqrt(x2 - 6x + 34)

I was trying to figure out the domain first, knowing that x2 - 6x >= -34 in order for the number inside the sqrt to be non-negative.. I don't know, maybe I'm missing something really basic here. Help? :)
 

Answers and Replies

  • #2
33,514
5,195

Homework Statement



Find the range of h

Homework Equations



h(x) = sqrt(25 + (x - 3)2)

The Attempt at a Solution



I factored out the (x-3)2 and simplified to get

sqrt(x2 - 6x + 34)
Let's get the terminology straight. You expanded (x-3)2 ; it was already factored.
I was trying to figure out the domain first, knowing that x2 - 6x >= -34 in order for the number inside the sqrt to be non-negative.. I don't know, maybe I'm missing something really basic here. Help? :)
The best thing, IMO, was to leave the radicand in its given form, 25 + (x-3)2. Looking at that as its own function, what is the range of this function? That will tell you a lot about the range of h(x).
 
  • #3
32
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Well the range of 25 + (x - 3)2 is y >= 25, right? So, the range of sqrt(25 + (x - 3)2 is y >= 5?

Yea, I think expanding the (x - 3)2 term messed me up. I was thinking that i was possible for the radicand to be a negative number.. ugh. Stupid mistake.

Thanks a lot for the help.
 
  • #4
Char. Limit
Gold Member
1,204
14
Also, just to let you know...

x^2-6x has a minimum of -9. It will never be less than -34.
 

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