[SOLVED] Functions

**unstopabl3** · October 7th 2009, 03:33 AM

Q) A function is defined by $f(x) = (2x-3)^3-8$ for x is greater than or equal to 2 and x is less than or equal to 4.

1) Find the range of f(x)
2) Find the domain of f(x)

Now normally I would use the completed square method to write the quadratic equations into a form such as $(x+b)^2+c$ and then plug in such a value for x to just leave +c behind which would be the range of my function and the value of x would become the domain. But as you can see the above equation is ^3 and not ^2. So what is the easiest method to find the range and the domain here?

Also another thing I wanted to ask is that in such questions when they say something like "for x is greater than or equal to 2 and x is less than or equal to 4" in front of the function, should we take this as the range or the domain or what? Please clarify a little bit on this thanks.

**mr fantastic** · October 7th 2009, 03:39 AM

Originally Posted by unstopabl3

Q) A function is defined by $f(x) = (2x-3)^3-8$ for x is greater than or equal to 2 and x is less than or equal to 4.

1) Find the range of f(x)
2) Find the domain of f(x)

Now normally I would use the completed square method to write the quadratic equations into a form such as $(x+b)^2+c$ and then plug in such a value for x to just leave +c behind which would be the range of my function and the value of x would become the domain. But as you can see the above equation is ^3 and not ^2. So what is the easiest method to find the range and the domain here?

Also another thing I wanted to ask is that in such questions when they say something like "for x is greater than or equal to 2 and x is less than or equal to 4" in front of the function, should we take this as the range or the domain or what? Please clarify a little bit on this thanks.

The domain is all real numbers. The range is all real numbers. See the range by drawing the graph (it's a cubic in the 'stationary point of inflection' form).

**unstopabl3** · October 7th 2009, 04:10 AM

Thanks for the quick reply but that just went over my head.

Isn't the range of f(x) = the domain of f^-1(x) and the domain of f(x) the range of f^-1(x) ???

According to that the range of f(x) for this question is x is greater than or equal to -7 and x is less than or equal to 117

Does that make sense?

I think you forgot to mention what "for x is greater than or equal to 2 and x is less than or equal to 4" means or symbolizes in the such function questions.

Thanks!

**Raoh** · October 7th 2009, 04:29 AM

Hello

you were right,the range of $f$ is the domain of $f^{-1}$ ,and the domain of $f^{-1}$ is the range of $f$ .

**unstopabl3** · October 7th 2009, 04:43 AM

Thanks for your reply, please answer the questions in my first post.

**Raoh** · October 7th 2009, 04:53 AM

there are some restrictions on $x$ , $\left \{ \ x\in \mathbb{R}\mid 2\leq x\leq 4 \right \}$
therefore the domain must be $[2,4]$ .
and the range is $[-7,117]$

**unstopabl3** · October 7th 2009, 06:05 AM

If that is the case shouldn't the domain be 2,3 and 4 instead of just 2 and 4 according to the inequality?

Also how did you figure out that range? Can you kindly solve and show?

**Raoh** · October 7th 2009, 06:14 AM

the domain is all real numbers between 2 and 4. $\left \{ \ x\in \mathbb{R}\mid 2\leq x\leq 4 \right \}$

**unstopabl3** · October 7th 2009, 06:18 AM

Same question as in #1 post but different part:

Find an expression, in terms of x, for $f^{-1}(x)$ and find the domain of $f^{-1}(x)$ .

Kindly solve this and show all of your steps as I'm unsure on how to solve this.

Thanks!

**unstopabl3** · October 7th 2009, 06:20 AM

Sorry double post.

**Raoh** · October 7th 2009, 06:57 AM

i can only give you the inverse of $f(x)=\left ( 2x-3 \right )^{3}$ $(\forall x\in \mathbb{R})$ (sorry

)
$y =\left ( 2x-3 \right )^{3}\Leftrightarrow \sqrt[3]{y}=2x-3\Leftrightarrow \sqrt[3]{y}+3=2x\Leftrightarrow \frac{\sqrt[3]{y}+3}{2}=x$
therefore the inverse is : $f^{-1}(y)= \frac{\sqrt[3]{y}+3}{2}$ .

**unstopabl3** · October 7th 2009, 07:56 AM

Thanks.

I have also solved till the the f^-1(x) part, I just don't know how to find out the domain and the range for this.

Let's wait and see if someone else can help.

**Raoh** · October 7th 2009, 11:33 AM

okay,it must be like this

$y+8=(2x-3)^{3}\Leftrightarrow \sqrt[3]{y+8}=2x-3\Leftrightarrow \sqrt[3]{y+8}+3=2x$ ,hence $x= \frac{\sqrt[3]{y+8}+3}{2}$ ,therefore $f^{-1}(y) = \frac{\sqrt[3]{y+8}+3}{2}$

**mr fantastic** · October 7th 2009, 05:50 PM

Originally Posted by unstopabl3

Thanks for the quick reply but that just went over my head.

Isn't the range of f(x) = the domain of f^-1(x) and the domain of f(x) the range of f^-1(x) ???

According to that the range of f(x) for this question is x is greater than or equal to -7 and x is less than or equal to 117

Does that make sense?

I think you forgot to mention what "for x is greater than or equal to 2 and x is less than or equal to 4" means or symbolizes in the such function questions.

Thanks!

Yes, I completely overlooked the restricted domain. However, I see the correct answer was subsequently given. (My comment about drawing the graph to get the range remains valid). As for domain and range of the inverse:

Originally Posted by unstopabl3

Thanks.
I have also solved till the the f^-1(x) part, I just don't know how to find out the domain and the range for this.
Let's wait and see if someone else can help.

what you said above (in red) in the first quote is exactly what you need to do.

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