What is the inverse of f(x)= 1/2 x+1/8 x^2 ?

The inverse of f(x)= 1/2 x+1/8 x^2 is 2(sqrt(2x+1)-1),-2(sqrt(2x+1)+1)

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inverse of f(x)= 1/2 x+1/8 x^2

inverse

f (x) = \frac{1}{2} x + \frac{1}{8} x^{2}

2 (2 x + 1 - 1), - 2 (2 x + 1 + 1)

Solution steps

y = \frac{1}{2} x + \frac{1}{8} x^{2}

Replace

x

with

y

x = \frac{1}{2} y + \frac{1}{8} y^{2}

Solve for

y, x = \frac{1}{2} y + \frac{1}{8} y^{2}

2 (2 x + 1 - 1), - 2 (2 x + 1 + 1)

5 sin (x) + 4

arcsec (\frac{1}{3} x)

f (x) = sin (x^{3})

f (x) = \frac{3 x}{1 - 2 x}

f (x) = (- 3.1)

What is the inverse of f(x)= 1/2 x+1/8 x^2 ?
The inverse of f(x)= 1/2 x+1/8 x^2 is 2(sqrt(2x+1)-1),-2(sqrt(2x+1)+1)