finding the g prime of 56

Question

If g(x) = x² + 10x with x ≥ −5, find g⁻¹(56).

Answer 1

g^-1 = -5 + sqr(x+25)

g^-1 = -5 + sqr(56+25)

=-5 + sqr81

=-5 +9

= 4

To get g^-1

let y = x² + 10x

then switch x and y and solve for y

x= y² + 10 y

add 25 and subtract 25, to complete the square

x = y² + 10 y + 25 - 25

factor

x = (y+5)² - 25

x+25 = (y+5)²

take the square root of both sides

y+5 = sqr(x+25)

subtract 5 from both sides

y= -5 + sqr(x+25) = g^-1(x)

ignore the negative square root as x>-5, never less than

g^-1(x) = -5 + sqr(x+25)

Answer 2

Find g⁻¹(56)

x²+10x-56=0

(x+14)(x-4)

x=-14 x=4

As x ≥ −5, x=4 would be your answer.