Range of a Function: Find h(x)

  • Thread starter Thread starter Codester09
  • Start date Start date
  • Tags Tags
    Function Range
AI Thread Summary
To find the range of the function h(x) = sqrt(25 + (x - 3)²), it's important to analyze the expression inside the square root. The term (x - 3)² is always non-negative, meaning the minimum value of the radicand is 25 when x = 3. Therefore, the range of h(x) is y ≥ 5, as the square root of 25 is 5. The discussion highlights that expanding the squared term can lead to confusion about the non-negativity of the radicand. Understanding the function's structure clarifies the range effectively.
Codester09
Messages
31
Reaction score
0

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? :)
 
Physics news on Phys.org
Codester09 said:

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.
Codester09 said:
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).
 
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.
 
Also, just to let you know...

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

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Find the domain and the range of ##f-3g##
Replies
3
Views
1K
To find the range of a given ##\sin## function
Replies
15
Views
2K
How to find the range of a function with square roots?
Replies
23
Views
2K
F(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?
Replies
4
Views
1K
Is it possible to find the range of this function?
Replies
22
Views
1K
Finding the domain (and range) of a logarithmic function
Replies
11
Views
3K
State the domain and range for a given function
Replies
7
Views
2K
Find the roots of the given cubic equation
Replies
13
Views
2K
Range of a function involving trigonometric functions
Replies
11
Views
2K
To find the boundedness of a given function
Replies
14
Views
3K

Hot Threads

Recent Insights

Back
Top