Tangent Bonanza

For (1):

Using the symmetry about the $y$ axis 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 the symmetry about the $y$ axis $\implies A^{'}: (\dfrac{1}{4^{\frac{1}{3}}}, \dfrac{-1}{4^{\frac{4}{3}}})$ and $B^{'}: (\dfrac{-1}{4^{\frac{4}{3}}},\dfrac{1}{4^{\frac{1}{3}}}) \implies$

$BB^{'} = \dfrac{2}{4^{\frac{4}{3}}}, AA^{'} = \dfrac{2}{4^{\frac{1}{3}}}$ and using point $P: (\dfrac{-1}{4^{\frac{4}{3}}},\dfrac{-1}{4^{\frac{4}{3}}})$ the height $h$ of the given trapezoid is $h = B^{'}P = \dfrac{5}{4^{\frac{4}{3}}} \implies A_{AB^{'}BA^{'}} = \boxed{(\dfrac{5}{4^{\frac{4}{3}}})^2}$

In addition:

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

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

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For (2):

Again using the symmetry about the $y$ axis 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 $m = 1$ and $a = -\dfrac{1}{2}$

$$

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 $D: (-\dfrac{1}{2},-\dfrac{1}{4})$

Using the symmetry about the $y$ axis $D:(\dfrac{-1}{2},\dfrac{-1}{4}) \rightarrow D^{'}:(\dfrac{1}{2},\dfrac{-1}{4})$ and $C:(\dfrac{1}{4},\dfrac{1}{2}) \rightarrow C^{'} :(\dfrac{-1}{4},\dfrac{1}{2})$ .

Using points $D^{'} :(\dfrac{1}{2},\dfrac{-1}{4})$ and $C^{'} :(\dfrac{-1}{4},\dfrac{1}{2}) \implies y = \dfrac{1}{4} - x$ and $DD^{'} = 1, CC^{'} = \dfrac{1}{2}$ and the height of the trapezoid $C^{'}Q = \dfrac{3}{4} \implies$ the area of the trapezoid $A_{DC^{'}CD^{'}} = \dfrac{1}{2}(\dfrac{3}{4})(\dfrac{3}{2}) = \dfrac{9}{16} = \boxed{(\dfrac{3}{4})^2}$

$\implies A_{AB^{'}BA^{'}} - A_{DC^{'}CD^{'}} = (\dfrac{5}{4^{\frac{4}{3}}})^2 - (\dfrac{3}{4})^2 \approx \boxed{0.0575785}$