Thread: Inverse functions

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

Thanks.

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.

$\displaystyle y = px + q$

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

You know that:

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

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

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

Could you please explain the steps to reach the answer

thanks.

Originally Posted by e^(i*pi)

$\displaystyle y = px + q$

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

You know that:

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

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

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

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

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

Rearrange eq2 to make q the subject:

Spoiler:

$\displaystyle -29-q = -2p$

$\displaystyle q = -29+2p$

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


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

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

$\displaystyle (35-2p) = 3p$

$\displaystyle p = 7$


Sub the value of p into (eq2):

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

$\displaystyle -(29+q) = -14$

$\displaystyle q = -15$

Test these values

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

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


Your equation now shows

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

Spoiler:

$\displaystyle f^{-1}(27) = \frac{1}{7}(27+15) = \frac{42}{7} = 6$

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