Functions

**gary223** · January 27th 2010, 03:12 AM

Hi all!

Just having a little problem with the following:

Functions f and g are defined as follows

$<br /> f(x) = 2x + 3, x\epsilon\Re$ and $g(x) = \frac{1}{x-1}, x\epsilon\Re$ , $x$ [does not equal to] 1

I couldn't find the correct symbol for not equal to sign.

(i) Find the expression for the inverse function f^-1 $(x)$

(ii) Find the expression for the composite function $gf$ $(x)$

(iii) Solve the equation f^-1 $(x)$ $=$ $gf(x)-1$

Thanks in advance.

**HallsofIvy** · January 27th 2010, 04:32 AM

Originally Posted by gary223

Hi all!

Just having a little problem with the following:

Functions f and g are defined as follows

$<br /> f(x) = 2x + 3, x\epsilon\Re$ and $g(x) = \frac{1}{x-1}, x\epsilon\Re$ , $x$ [does not equal to] 1

I couldn't find the correct symbol for not equal to sign.

$\ne$ "\ne".

(i) Find the expression for the inverse function f^-1 $(x)$

If y= f(x), then $x= f^{-1}(y)$ . If y= 2x+3, solve for x, then "swap" x and y so you can write it as $f^{-1}(x)$ rather than $f^{-1}(y)$ .

(ii) Find the expression for the composite function $gf$ $(x)$

g(f(x))= g(2x+3). Replace "x" by "2x+3" in the formula for g.

(iii) Solve the equation f^-1 $(x)$ $=$ $gf(x)-1$

Thanks in advance.

Use your answers to (i) and (ii) to get the equation for (iii).

**gary223** · January 27th 2010, 05:20 AM

Originally Posted by HallsofIvy

$\ne$ "\ne".

If y= f(x), then $x= f^{-1}(y)$ . If y= 2x+3, solve for x, then "swap" x and y so you can write it as $f^{-1}(x)$ rather than $f^{-1}(y)$ .

g(f(x))= g(2x+3). Replace "x" by "2x+3" in the formula for g.

Use your answers to (i) and (ii) to get the equation for (iii).

Thanks for the helpful reply but I'm still having problems actually solving the three parts. Could you post the answer so I can work from that example when trying to do other questions I have similar to this.

**HallsofIvy** · January 27th 2010, 06:16 AM

Originally Posted by gary223

Thanks for the helpful reply but I'm still having problems actually solving the three parts. Could you post the answer so I can work from that example when trying to do other questions I have similar to this.

How about you post what you have done?

In the first problem, I suggested that you solve y= 2x+3 for x. Did you do that?

In the second, I told you to replace x by 2x+ 3 in the formula for g(x). Since that formula was $g(x)= \frac{1}}{x-1}$ , replacing x by 2x-3 gives $g(f(x))= g(2x+3)= \frac{1}{(2x+3)-1}$ . What is that?

**gary223** · January 27th 2010, 10:15 AM

Here's what I got for the first part:

$y = 2x+3$
$y - 3 = 2x$
$\frac{y-3}{2} = x$

Correct?

**gary223** · January 28th 2010, 07:47 AM

And I got $g(f(x)) = \frac{1}{2x+2}$

Please help.

**HallsofIvy** · January 29th 2010, 01:36 AM

Originally Posted by gary223

Here's what I got for the first part:

$y = 2x+3$
$y - 3 = 2x$
$\frac{y-3}{2} = x$

Correct?

Very good- so far! Now answer the question- what is $f^{-1}(x)$ ?

**HallsofIvy** · January 29th 2010, 01:37 AM

Originally Posted by gary223

And I got $g(f(x)) = \frac{1}{2x+2}$

Please help.

Yes, that is correct.

**gary223** · January 29th 2010, 06:11 AM

For iii)

I got:

$\frac{y-3}{2} = \frac{1}{2x+2} - 1$

$=\frac{y-3+1}{2} = \frac{1}{2x+2}$

$=\frac{y-2}{2} = \frac{1}{2x+2}$

Is this correct? If not, what is the rest?

**HallsofIvy** · January 30th 2010, 02:09 AM

Originally Posted by gary223

Here's what I got for the first part:

$y = 2x+3$
$y - 3 = 2x$
$\frac{y-3}{2} = x$

Correct?

Originally Posted by HallsofIvy

Very good- so far! Now answer the question- what is $f^{-1}(x)$ ?

Originally Posted by gary223

For iii)

I got:

$\frac{y-3}{2} = \frac{1}{2x+2} - 1$

$=\frac{y-3+1}{2} = \frac{1}{2x+2}$

$=\frac{y-2}{2} = \frac{1}{2x+2}$

Is this correct? If not, what is the rest?

I told you before that you had not finished problem 1. You got $\frac{y-3}{2} = x$ But the problem was to find $f^{-1}(x)$ , not $f^{-1}(y)$ ! You need to write "x" in place of "y" to make that a function of x.

That's why you have an equation involving both x and y when you should have an equation in x only. Problem iii was to solve $f^{-1}(x)= f(g(x))- 1$ , an equation in x only, for x.

**gary223** · February 1st 2010, 05:01 AM

Originally Posted by HallsofIvy

I told you before that you had not finished problem 1. You got $\frac{y-3}{2} = x$ But the problem was to find $f^{-1}(x)$ , not $f^{-1}(y)$ ! You need to write "x" in place of "y" to make that a function of x.

That's why you have an equation involving both x and y when you should have an equation in x only. Problem iii was to solve $f^{-1}(x)= f(g(x))- 1$ , an equation in x only, for x.

So its $\frac{x-3}{2} = y$ ?

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