Range of a Function: Find h(x)

[Codester09]

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? :)

 

[Mark44]

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).

 

[Codester09]

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.

 

[Char. Limit]

Also, just to let you know...

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