Question

A function of f is f(x) = px + q. Given that f(1) = -5 and f(-2) = -10, find the value of p and q. Hence, find f⁻¹(x).

Answer 1

Answer:

$p = \frac{5}{3} \\ \\ q = - 5 - \frac{5}{ 3} \\ \\ q = \frac{ - 20}{3} \\ \\ f(x) = \frac{5x}{3} - \frac{20}{3} \\ \\therefore \: \: \: {f}^{ - 1} (x) = \frac{3x + 20}{5} .$

Step-by-step explanation:

$f(x) = px + q \\ \\ f(1) = - 5 \\ \\ p + q = - 5 \: \: \: \: ...(i) \\ \\ f(-2) = - 2p + q = - 10 \: \: \: \: ...(ii) \\ \\ solving \: these \: \: we \: get \\ \\ 3p = 5 \\ \\ p = \frac{5}{3} \\ \\ q = - 5 - \frac{5}{ 3} \\ \\ q = \frac{ - 20}{3} \\ \\ f(x) = \frac{5x}{3} - \frac{20}{3} \\ \\ f(x) = \frac{5(x - 4)}{3} \\ \\ therefore \: \: \: \: 3y = 5x - 20 \\ \\ 5x = 3y + 20 \\ \\ x= \frac{3y + 20}{5} \\ \\ therefore \: \: \: {f}^{ - 1} (x) = \frac{3x + 20}{5} .$

Answer 2

Answer:

given f(x)=px+q & f(1)=-5,f(-2)=-10

f(1)=p(1)+q&f(-2)=p(-2)+q

-5=p+q&-10=-2p+q

by solving these two

p+q=-5

-2p+q=-10

now we get 3p=5

p= 5/3

5/3+q=-5

q=-5-5/3

q=-15-5/3

q=-10/3

let f(x)=y

x= f inverse of y

px+q=y

5/3-10/3=y

-5=y

swap x and y

f(y)=x

y= f inverse of x

so f inverse of x = -5