How do you find the inverse of #g(x) = x^2 + 4x + 3 # and is it a function?

Question

How do you find the inverse of  #g(x) = x^2 + 4x + 3 # and is it a function?

Elijah Allen · Accepted Answer

See below.

To find the inverse function, we need to express #x# as a function of #y#: #y=x^2+4x+3# Replace: #y=x# #x=y^2+4y+3# Subtract #x#: #y^2+4y+3-x=0# Using the formula for quadratics: #y=(-(4)+-sqrt((4)^2-4(1)(3-x)))/(2(1))# #y=(-4+-sqrt(16-12+4x))/2# #y=(-4+-sqrt(4+4x))/2# #y=(-4+-sqrt(4(1+x)))/2# #y=(-4+-2sqrt((1+x)))/2=-2+-sqrt((1+x))# #:.# #f^-1(x)=-2+sqrt(1+x)# #f^-1(x)=-2-sqrt(1+x)# If we look at the inverses, remembering that #x# is the range of the function, we can see that, for: #sqrt(1+x)# #1+x>=0# #x>=-1# This implies that: #x^2+4x+3>=-1# Solving: #x^2+4x+4>=0# Factor: #(x+2)^2>=0# Thus, the inverses are defined for the function's domain. They are both functions if we take #sqrt(1+x)# as meaning the principal root.

Ariana Alvarez · Answer

undefined To find the inverse of a function, you first need to replace the function notation with $ y $. Then, swap $ x $ and $ y $ and solve for $ y $. Finally, replace $ y $ with $ f^{-1}(x) $ to express the inverse function.

Given $ g(x) = x^2 + 4x + 3 $:

1. Replace $ g(x) $ with $ y $: $ y = x^2 + 4x + 3 $
2. Swap $ x $ and $ y $: $ x = y^2 + 4y + 3 $
3. Solve for $ y $:
   $$ x = y^2 + 4y + 3 $$
   $$ 0 = y^2 + 4y + 3 - x $$
   $$ y^2 + 4y + (3 - x) = 0 $$
   $$ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$
   $$ y = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(3 - x)}}}}{{2(1)}} $$
   $$ y = \frac{{-4 \pm \sqrt{{16 - 4(3 - x)}}}}{2} $$
   $$ y = \frac{{-4 \pm \sqrt{{16 - 12 + 4x}}}}{2} $$
   $$ y = \frac{{-4 \pm \sqrt{{4 + 4x}}}}{2} $$
   $$ y = \frac{{-4 \pm 2\sqrt{{1 + x}}}}{2} $$
   $$ y = -2 \pm \sqrt{{1 + x}} $$
4. Replace $ y $ with $ g^{-1}(x) $:
   $$ g^{-1}(x) = -2 \pm \sqrt{{1 + x}} $$

The inverse function of $ g(x) = x^2 + 4x + 3 $ is $ g^{-1}(x) = -2 \pm \sqrt{{1 + x}} $.

To determine if it's a function, we need to check if each input $ x $ corresponds to exactly one output $ y $. Since the inverse function is given as $ g^{-1}(x) = -2 \pm \sqrt{{1 + x}} $, it has two possible outputs for each input $ x $. Therefore, $ g^{-1}(x) $ is not a function.