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 $\displaystyle f^{-1}(27)$
$\displaystyle \frac{1}{p}(6-q) = 3 $...............(eq1)
$\displaystyle \frac{1}{p}(-29-q) = -2$.............(eq2)
Rearrange eq2 to make q the subject:
Spoiler:
$\displaystyle q = -29+2p$.........................(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: