Tangent Problem 2

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 $h = B^{'}P = \dfrac{5}{4^{\frac{4}{3}}} \implies A_{AB^{'}BA^{'}} = (\dfrac{5}{4^{\frac{4}{3}}})^2 =$ $(\dfrac{5}{2^2 * 2^{\frac{2}{3}}})^2 = (\dfrac{\lambda}{\alpha^{\alpha} * \alpha^{\frac{\alpha}{\beta}}})^{\alpha} \implies \alpha + \beta + \lambda = \boxed{10}$ .

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In addition:

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

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