What's the Distance?

Let ${ f(x) = -x^4 }$ and ${ g(x) = x^{1/4} \implies }$

$\dfrac{d}{dx}(f(x))|_{x = a} = -4a^3$ and $\dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{4b^{\frac{3}{4}}} \implies$ $-4a^3 = \dfrac{1}{4b^{\frac{3}{4}}} \implies a = \dfrac{-1}{16^{\frac{1}{3}}b^\frac{1}{4}}$

Using $A: (a,-a^4) = (\dfrac{-1}{16^{\frac{1}{3}}b^\frac{1}{4}},\dfrac{-1}{16^{\frac{4}{3}}b})$ and $B: (b,b^{\frac{1}{4}}) \implies$

The slope $m = \dfrac{16^{\frac{4}{3}} b^{\frac{5}{4}} + 1}{(16b^{\frac{3}{4}})(16^{\frac{1}{3}} b^{\frac{5}{4}} + 1)} = \dfrac{1}{4b^{\frac{3}{4}}} \implies$ $4^{\frac{8}{3}}b^{\frac{5}{4}} + 1 = 4^{\frac{5}{3}}b^{\frac{5}{4}} + 4 \implies$ $4^{\frac{5}{3}}b^{\frac{5}{4}} = 1 \implies b = \dfrac{1}{4^{\frac{4}{3}}} \implies a = \dfrac{-1}{4^{\frac{1}{3}}}$

$\implies A: (\dfrac{-1}{4^{\frac{1}{3}}}, \dfrac{-1}{4^{\frac{4}{3}}})$ and $B: (\dfrac{1}{4^{\frac{4}{3}}},\dfrac{1}{4^{\frac{1}{3}}})$

Using points $A$ and $B \implies m_{AB} = 1 \implies y = x + \dfrac{3}{4^{\frac{4}{3}}}$

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Now let ${ f(x) = -x^2 }$ and ${ g(x) = \sqrt{x} \implies }$

$\dfrac{d}{dx}(f(x))|_{x = a} = -2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{2 \sqrt{b}} \implies$

$-2a = \dfrac{1}{2 \sqrt{b}} \implies a = -\dfrac{1}{4 \sqrt{b}}$

$D: (a,-a^2) = (-\dfrac{1}{4 \sqrt{b}} , -\dfrac{1}{16 b})$ and $C: (b, \sqrt{b} ) \implies$

slope $m = \dfrac{1}{2 \sqrt{b}} = \dfrac{1}{4 \sqrt{b}} (\dfrac{16 b^\frac{3}{2} + 1}{4 b^\frac{3}{2} + 1}) \implies$

$16 b^\frac{3}{2} + 1 = 8 b^\frac{3}{2} + 2 \implies 8 b^\frac{3}{2} = 1 \implies b = \dfrac{1}{4}$

$\implies$ slope $m_{DC} = 1$ and $a = -\dfrac{1}{2}$

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Using $D: (-\dfrac{1}{2},-\dfrac{1}{4}) \implies y = x + \dfrac{1}{4}$

$y = x + \dfrac{1}{4}$ is tangent to $g(x)$ at $C: (\dfrac{1}{4} , \dfrac{1}{2})$ and $f(x)$ at $D: (-\dfrac{1}{2},-\dfrac{1}{4})$

To find the perpendicular distance between $\overleftrightarrow{AB}$ and $\overleftrightarrow{AB}$ .

For $\overleftrightarrow{AB}$ , $y = x + \dfrac{3}{4^{\frac{4}{3}}}$

For $\overleftrightarrow{CD}$ , $y = x + \dfrac{1}{4}$

Let $\overleftrightarrow{DF}$ be the line perpendicular to $y = x + \dfrac{1}{4}$ and $y = x + \dfrac{3}{4^{\frac{4}{3}}}$

$\overleftrightarrow{DF}$ pass thru the point $(\dfrac{-1}{2},\dfrac{-1}{4})$ with slope $m_{\perp} = -1 \implies$

$y = -x - \dfrac{3}{4}$ represents line $\overleftrightarrow{DF}$

Find the point of intersection for the lines:

$y = x + \dfrac{3}{4^{\frac{4}{3}}}$

$y = -x - \dfrac{3}{4}$

$\implies x = \dfrac{-3}{8}(1 + \dfrac{1}{4^{\frac{1}{3}}})$ and $y = \dfrac{-3}{8}(1 - \dfrac{1}{4^{\frac{1}{3}}})$

$\implies$ the distance $DF = \dfrac{1}{8}\sqrt{\dfrac{4^{\frac{4}{3}} + 9 * 4^{\frac{2}{3}} - 24}{2 * 4^{\frac{1}{3}}}} \approx \boxed{.157310}$ after simplifying.