How do I find the inverse of the following notation?

Question

I am having trouble finding the inverse of the function notation:

f(x)=x^2 + 4x - 3

Please help! I tried one way and it didn't work but I dunno what else to do!

Answer 1

First, use the technique of completing the square to rewrite f(x):

x^2+bx+c can be written as

(x+b/2)^2 + c - b^2/4.

In this case, b=4 and c=-3, so you have

f(x)=(x+2)^2-7

To find the inverse: set y=(x+2)^2-7.

Then solve for x:

y+7=(x+2)^2

sqrt(y+7)-2=x
or
-sqrt(y+7)-2=x

Interchange x and y to obtain the possible inverse functions:

f^-1(x) = sqrt(x+7)-2
or
f^-1(x) = -sqrt(x+7)-2.

f(x) does not have an inverse defined on the entire real number line, because it is not a one-to-one function. However, if you restrict the domain of f(x) to x >= -2, or x <= -2, then it does have an inverse (choose the first one for x >= -2, or the second one for x <= -2)

Answer 2

well Darling ;

The original function:
f(x)= x^2 + 4x - 3

Rename "f(x)" as "y":
y = x^2 + 4x - 3

0 = x^2 + 4x - 3 -y
0 = x^2 + 4x - (3 -y)
Solve for "x =" by using the Quadratic Formula:
x = - (4) + √(4^2) - 4(1)(3-y) / 2(1) = -4 + √(8-12 +4y)/2 =
-4 + √(-4 +4y)/2
x = - (4) - √(4^2) - 4(1)(3-y) / 2(1) = -4 - √(8-12 +4y)/2
-4 - √(-4 +4y)/2

Since x ≤-2, then we want the negative square root:
-4 + √(-4 +4y)/2

Switch x and y:
y = -4+ √(-4 +4x)/2

Rename "y" as "f-inverse"; the domain restriction comes from the fact that this is a rational function.
f^-1(x) = -4+ √(-4 +4x)/2

Then the inverse is f^-1(x) = -4+ √(-4 +4x)/2
(-4 +4x) ≥ 0
4x ≥ +4
x ≥ 1
Domain is x ≥ 1 OR [+1 , + ∞)

Good Luck Darling.

Answer 3

The way to find the inverse of a function is to change all the "x" to "y", and all the "y" (or f(x) in your case) to "x", then re-solve for y:

f(x)=x^2 + 4x - 3
same thing as y=x^2 + 4x -3

change it around...

x=y^2 + 4y - 3

factor by completing the square...
x+7=y^2 + 4y + 4
x+7=(y+2)^2

square root both sides...
+/-sqrt(x+7)=y+2

y=2+/- sqrt(x+7)

Answer 4

Solve for x by completing the square.