# Square Root Function

• May 17th 2008, 10:00 AM
chrozer
Square Root Function
Create a square root function with:

(a) domain of $\displaystyle [3, \infty)$ and a range of $\displaystyle (-\infty, 5]$.

Would this be right? $\displaystyle f(x)=-\sqrt{x+3}+5$

And how would you create a square root function that has a domain of all real numbers?
• May 17th 2008, 10:06 AM
Moo
Hello,

Quote:

Originally Posted by chrozer
Create a square root function with:

(a) domain of $\displaystyle [3, \infty)$ and a range of $\displaystyle (-\infty, 5]$.

Would this be right? $\displaystyle f(x)=-\sqrt{x+3}+5$

It would be better if it was $\displaystyle f(x)=-\sqrt{x{\color{red}-}3}+5$
(Wink)

Quote:

And how would you create a square root function that has a domain of all real numbers?
$\displaystyle \sqrt{x^2+\dots}$

:p
• May 17th 2008, 10:15 AM
chrozer
Quote:

Originally Posted by Moo
Hello,

It would be better if it was $\displaystyle f(x)=-\sqrt{x{\color{red}-}3}+5$
(Wink)

$\displaystyle \sqrt{x^2+\dots}$

:p

Duh....that was a stupid mistake. And thanks for the second part.

Two more things...how would you determine the domain and range of $\displaystyle f(x)=\sqrt{ax+b}+c$ in terms of $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$.

And would $\displaystyle f^{-1}(x)=\frac {(x-c)^2-b}{a}$ be the inverse of $\displaystyle f(x)=\sqrt{ax+b}+c$?
• May 17th 2008, 10:25 AM
Mathstud28
Quote:

Originally Posted by chrozer
Duh....that was a stupid mistake. And thanks for the second part.

Two more things...how would you determine the domain and range of $\displaystyle f(x)=\sqrt{ax+b}+c$ in terms of $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$.

And would $\displaystyle f^{-1}(x)=\frac {(x-c)^2-b}{a}$ be the inverse of $\displaystyle f(x)=\sqrt{ax+b}+c$?

For $\displaystyle \sqrt{ax+b}+c$ to be $\displaystyle \in\mathbb{R}$

$\displaystyle ax+b\geq{0}\Rightarrow{x\geq\frac{-b}{a}}$

and for the second one since $\displaystyle f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}$

Now just solve for $\displaystyle f^{-1}(x)$
• May 17th 2008, 10:31 AM
chrozer
Quote:

Originally Posted by Mathstud28
For $\displaystyle \sqrt{ax+b}+c$ to be $\displaystyle \in\mathbb{R}$

$\displaystyle ax+b>0\Rightarrow{x>\frac{-b}{a}}$

and for the second one since $\displaystyle f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}$

Now just solve for $\displaystyle f^{-1}(x)$

Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero?
• May 17th 2008, 10:35 AM
Moo
Quote:

Originally Posted by chrozer
Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero?

For the range :

You know that $\displaystyle \sqrt{ax+b}$ goes from 0 to $\displaystyle +\infty$

So $\displaystyle \sqrt{ax+b}+c$ goes from 0+c=c to $\displaystyle +\infty$

:)
• May 17th 2008, 10:43 AM
chrozer
Quote:

Originally Posted by Moo
For the range :

You know that $\displaystyle \sqrt{ax+b}$ goes from 0 to $\displaystyle +\infty$

So $\displaystyle \sqrt{ax+b}+c$ goes from 0+c=c to $\displaystyle +\infty$

:)

So all together the domain would be $\displaystyle x \geq \frac{-b}{a}$ and the range would be $\displaystyle y \geq c$. Right?
• May 17th 2008, 10:44 AM
Moo
Quote:

Originally Posted by chrozer
So all together the domain would be $\displaystyle x \geq \frac{-b}{a}$ and the range would be $\displaystyle c \leq y \leq +\infty$. Right?

Correct for the domain.
For the range, it's like $\displaystyle c \leq y$ (Wink)
• May 17th 2008, 10:50 AM
chrozer
Quote:

Originally Posted by Moo
Correct for the domain.
For the range, it's like $\displaystyle c \leq y$ (Wink)

Oh I see. Thnx alot.