Inverse functions

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October 27th 2009, 11:33 AM
Oasis1993

Inverse functions

The function f(x)= px+q such that f-1(6)=3 and f-1(-29)= -2. Find f-1(27).

Thanks.
October 27th 2009, 12:04 PM
e^(i*pi)

Quote:

Originally Posted by Oasis1993

The function f(x)= px+q such that f-1(6)=3 and f-1(-29)= -2. Find f-1(27).

Thanks.

$y = px + q$

$x = \frac{1}{p}(y-q)$ . Therefore $f^{-1}(x) = \frac{1}{p}(x-q)$

You know that:

$f^{-1}(6) = \frac{1}{p}(6-q) = 3$

$f^{-1}(-29) = \frac{1}{p}(-29-q) = -2$

Solve for p and q then sub those values to find $f^{-1}(27)$
October 27th 2009, 12:39 PM
Oasis1993

Could you please explain the steps to reach the answer

thanks.
October 27th 2009, 01:14 PM
e^(i*pi)

Quote:

Originally Posted by e^(i*pi)

$y = px + q$

$x = \frac{1}{p}(y-q)$ . Therefore $f^{-1}(x) = \frac{1}{p}(x-q)$

You know that:

$f^{-1}(6) = \frac{1}{p}(6-q) = 3$

$f^{-1}(-29) = \frac{1}{p}(-29-q) = -2$

Solve for p and q then sub those values to find $f^{-1}(27)$

$\frac{1}{p}(6-q) = 3$ ...............(eq1)

$\frac{1}{p}(-29-q) = -2$ .............(eq2)

Rearrange eq2 to make q the subject:

Spoiler:

$-29-q = -2p$

$q = -29+2p$

$q = -29+2p$ .........................(eq 2.1)

Sub the RHS of (eq 2.1) wherever q appears in (eq1):

$\frac{1}{p}(6-(-29+2p)) = 3$

$(35-2p) = 3p$

$p = 7$

Sub the value of p into (eq2):

$\frac{1}{7}(-29-q) = -2$

$-(29+q) = -14$

$q = -15$

Test these values

$f^{-1}(6) = \frac{1}{7}(6-(-15)) = 3$ . This is correct

$f^{-1}(-29) = \frac{1}{7}(-29-(-15)) = -2$ . This is correct

Your equation now shows

$f^{-1}(x) = \frac{1}{7}(x+15)$ so you should be able to find $f^{-1}(27)$ by direct substitution.

Spoiler:

$f^{-1}(27) = \frac{1}{7}(27+15) = \frac{42}{7} = 6$
October 28th 2009, 07:14 AM
Oasis1993

Thank you very much for the help!
it helped alot!!!!!!!