calculus - exponents and logarithms

Question

Find the value of b in the function f(x)=ax+b if f−1(2)=3 and f−1(−3)=6, where f−1(x) is the inverse of the function f(x). b=

Answer 1

b = 7

f(x) = ax + b

f(6) = a(6) + b = -3 when f^-1(-3) =6

f(3) = a(3) +b = 2 when f^-1(2) = 3

subtract the equations to get

3a = -5

a = -5/3

3a + b =2

3(-5/3) + b = 2

-5 + b =2

b = 2+5 = 7 which is the answer to the problem, b=7

f(x) = -5x/3 + 7

to find the inverse function switch x and y and solve for the new y which is f^-1(x)

y=-5x/3 + 7

x =-5y/3 + 7

5y/3 = -x + 7

y = -3x/5 + 21/5 = f^-1(x)

f^-1(x) = -3x/5 + 21/5

f^-1(2) = -3(2)/5 +21/5 =(21-6)/5 = 15/5 = 3

f^-1(2) = -3(-3)/5 + 21/5 = (21+9)/5 = 30/5 = 6

(not sure why this problem is listed under calculus, exponents and logs? It just involves two linear equations or functions, graphically two straight lines, both downward sloping. Just basic algebra.)