Lesson Finding Inverse

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This Lesson (Finding Inverse) was created by by Nate(3500) About Me : View Source, Show
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Inverses of graphs are about finding symmetrical graphed lines along the identity function.
To find these inverse graphed lines, you need to switch the variables.
Example:
f(x) = 2x + 1
y = 2x + 1
x = 2y + 1
x - 1 = 2y
0.5x - 0.5 = y
0.5x - 0.5 = f^-1(x)
graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+2x+%2B+1%2C+0.5x+-+0.5%2C+x+%29+
If you see a patter for the points of the lines, you can see that the values are switched as well.
Points for f(x): (0,1), (1,3), and (2,5)
Points for f^-1(x): (1,0), (3,1), and (5,2)
Example:
f(x) = 1/x - 1
y = 1/x - 1
x = 1/y - 1
x + 1 = 1/y
y(x + 1) = 1
y = 1/(x + 1)
f^-1(x) = 1/(x + 1)
graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+1%2Fx+-+1%2C+1%2F%28x+%2B+1%29%2C+x+%29+
Example:
f(x) = x^2 - 1
y = x^2 - 1
x = y^2 - 1
x + 1 = y^2
+-sqrt(x + 1) = y
+-sqrt(x + 1) = f^-1(x)
graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2+-+1%2C+sqrt%28x+%2B+1%29%2C+-sqrt%28x+%2B+1%29%2C+x+%29+
Example:
f(x) = 1/(x^2 + 2)
y = 1/(x^2 + 2)
x = 1/(y^2 + 2)
(y^2 + 2)x = 1
y^2 + 2 = 1/x
y^2 = 1/x - 2
y = +-sqrt(1/x - 2)
f^-1(x) = +-sqrt(1/x - 2)
graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+1%2F%28x%5E2+%2B+2%29%2C+sqrt%281%2Fx+-+2%29%2C+-sqrt%281%2Fx+-+2%29%2C+x+%29+
Example:
f(x) = sqrt(x^2) - 2
y = sqrt(x^2) - 2
x = sqrt(y^2) - 2
x + 2 = sqrt(y^2)
(x + 2)^2 = y^2
+-sqrt((x + 2)^2) = f^-1(x)
graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%28x%5E2%29+%2B+2%2C+sqrt%28%28x+-+2%29%5E2%29%2C+-sqrt%28%28x+-+2%29%5E2%29%2C+x+%29+

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