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]
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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].
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