Expert: Sherman D. - 4/27/2007

Question
Ok I'm having trouble turning these equations algebraically to inverse funtions....I think I know how to do it, I'm just not too sure....Please help, thanks.

1/2x^2+2

(x+3)^3+4

(2x+4)/x+3

and I also need help simplyfying these exponential functions.

(-2^4*16^-1)/8^2

((6x^-1y^4)/3xy)

And solve....
9^x(3^7x-1)=3^3+x

Answer
y = (1/2)x^2 + 2
x = (1/2)y^2 + 2
x - 2 = (1/2)y^2
2x - 4 = y^2
y = sqrt(2x - 4)
ANS : y^-1(x) = sqrt(2x - 4)

unless you meant

1/(2x^2 + 2)

y = 1/(2x^2 + 2)
y = 1/(2(x^2 + 1))
x = 1/(2(y^2 + 1))
2x = 1/(y^2 + 1)
1/(2x) = y^2 + 1
(1/(2x)) - 1 = y^2
y^2 = ((1 - 2x)/(2x))
y = sqrt((1 - 2x)/(2x))
y = (sqrt(2x - 4x^2))/(2x)
ANS : y^-1(x) = (sqrt(2x - 4x^2))/(2x)

------------------------------------

(x + 4)^3 + 4

y = (x + 4)^3 + 4
x = (y + 4)^3 + 4
x - 4 = (y + 4)^3
cbrt(x - 4) = y + 4
y = cbrt(x - 4) - 4
ANS : y^-1(x) = cbrt(x - 4) - 4

----------------------------------------------------

(2x + 4)/(x + 3) or ((2x + 4)/x) + 3

y = (2x + 4)/(x + 3)
x = (2y + 4)/(y + 3)
xy + 3x = 2y + 4
xy - 2y = -3x + 4
y(x - 2) = -3x + 4
y = (-3x + 4)/(x - 2)
ANS : y^-1(x) = (-3x + 4)/(x - 2)

Unless you meant

((2x + 4)/x) + 3

y = ((2x + 4)/x) + 3
x = ((2y + 4)/y) + 3
x - 3 = (2y + 4)/y
xy - 3y = 2y + 4
xy - 5y = 4
y(x - 5) = 4
y = 4/(x - 5)
ANS : y^-1(x) = 4/(x - 5)

Just switch x and y and solve for y.

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If by this you meant

((-2)^4 * 16^(-1))/(8^2)
(16 * (1/16))/64
1/64

Unless you meant

(-(2^4) * 16^(-1))/(8^2)
(-16 * (1/16))/64
-1/64

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(6x^-1 * y^4)/(3xy)
(6/3) * x^(-1 - 1) * y^(4 - 1)
2 * x^(-1 + (-1)) * y^3
ANS : (2y^3)/(x^2)

(9^x)(3^(7x - 1)) = 3^(3 + x)
((3^2)^x)(3^(7x - 1)) = 3^(3 + x)
(3^(2x))(3^(7x - 1)) = 3^(3 + x)
3^(2x + 7x - 1) = 3^(3 + x)
3^(9x - 1) = 3^(3 + x)
9x - 1 = 3 + x
8x = 4
x = (1/2)