# Determine the inverse function of f(x)=2x/2(x^3+x) .

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We have f(x) = 2x / 2(x^3 + x)

Let y = f(x) = 2x / 2(x^3 + x)

=> y = 2x / 2x(x^2 +1)

cancel 2x

=> y = 1/ x^2 + 1)

=> x^2 + 1 = 1/y

=> x^2 = 1/ y - 1

=> x^ 2 = (1 - y)/y

=> x = sqrt [ (1 - y)/y]

exchange x and y

=> y = sqrt [ (1 - x)/x]

The inverse function is

**f(x) = sqrt [ (1 - x)/x]**

For the beginning, we'll simplify the fraction that represents the expression of the function. We'll factorize by x the denominator:

f(x) = 2x/2x(x^2 + 1)

We'll simplify and we'll get:

f(x) = 1/(x^2 + 1)

We'll write f(x) = y.

y = 1/(x^2 + 1)

We'll change y by x:

x = 1/(y^2 + 1)

We'll multiply by 1/(y^2 + 1) both sideS:

x(y^2 + 1) = 1

We'll remove the brackets:

xy^2 + x = 1

We'll subtract x both sides:

xy^2 = 1 - x

We'll divide by x:

y^2 = (1-x)/x

We'll take radicals both sides:

sqrt y^2 = sqrt[(1-x)/x]

y = sqrt[(1-x)/x]

**The inverse function is f^-1(x) = sqrt[(1-x)/x].**