What is the inverse function of $f (x) = 2 x + x^{\frac{1}{2}}$ $f(x) = 2x + x^(1/2)$ ?

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Answer 1 · 2018-07-10T13:03:10

$y = \frac{4 x + 1 + \sqrt{8 x + 1}}{8}$ $color(blue)(y=(4x+1+sqrt(8x+1))/8$

$y = \frac{4 x + 1 - \sqrt{8 x + 1}}{8}$ $color(blue)(y=(4x+1-sqrt(8x+1))/8$

Exchange $x$ $x$ and $y$ $y$ :

$x = 2 y + y^{\frac{1}{2}}$ $x=2y+y^(1/2)$

Subtract $2 y$ $2y$ :

$x - 2 y = y^{\frac{1}{2}}$ $x-2y=y^(1/2)$

Square both sides:

${(x - 2 y)}^{2} = y$ $(x-2y)^2=y$

Expand $L H S$ $LHS$

$x^{2} - 4 x y + 4 y^{2} = y$ $x^2-4xy+4y^2=y$

$x^{2} - 4 x y + 4 y^{2} - y = 0$ $x^2-4xy+4y^2-y=0$

$4 y^{2} - (4 x + 1) y + x^{2} = 0$ $4y^2-(4x+1)y+x^2=0$

We now solve this for $y$ $y$ , using the quadratic formula:

$y = \frac{- (- 4 x - 1) \pm \sqrt{{(- 4 x - 1)}^{2} - 4 (4) (x^{2})}}{2 (4)}$ $y=(-(-4x-1)+-sqrt((-4x-1)^2-4(4)(x^2)))/(2(4)$

$y = \frac{- (- 4 x - 1) + \sqrt{{(- 4 x - 1)}^{2} - 4 (4) (x^{2})}}{2 (4)}$ $y=(-(-4x-1)+sqrt((-4x-1)^2-4(4)(x^2)))/(2(4)$

$y = \frac{4 x + 1 + \sqrt{8 x + 1}}{8}$ $color(blue)(y=(4x+1+sqrt(8x+1))/8$

$y = \frac{- (- 4 x - 1) - \sqrt{{(- 4 x - 1)}^{2} - 4 (4) (x^{2})}}{2 (4)}$ $y=(-(-4x-1)-sqrt((-4x-1)^2-4(4)(x^2)))/(2(4)$

$y = \frac{4 x + 1 - \sqrt{8 x + 1}}{8}$ $color(blue)(y=(4x+1-sqrt(8x+1))/8$

What is the inverse function of f(x)=2x+x12f(x) = 2x + x^(1/2)?