Area of a Region

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

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

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

${\bf A: (a,-a^2) = (-\frac{1}{4 \sqrt{b}} , -\frac{1}{16 b}) }$ ${\bf B: (b, \sqrt{b} ) \implies }$

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

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

${\bf \implies }$ slope ${\bf m = 1 }$ and ${\bf a = -\frac{1}{2} }$

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Using ${\bf A: (-\frac{1}{2},-\frac{1}{4}) \implies y = x + \frac{1}{4} }$

${\bf y = x + \frac{1}{4} }$ is tangent to ${\bf g(x)}$ at ${\bf B: (\frac{1}{4} , \frac{1}{2}) }$ and ${\bf f(x) }$ at ${\bf A: (-\frac{1}{2},-\frac{1}{4}) }$

${\bf \therefore }$ The desired area is A = ${\bf \int_{-\frac{1}{2}}^{0} (x + \frac{1}{4} + x^2) dx + \int_{0}^{\frac{1}{4}} (x + \frac{1}{4} - x^{\frac{1}{2}}) dx = }$

${\bf \frac{1}{4} * ( \int_{-\frac{1}{2}}^{0} (4x^2 + 4x + 1) dx + \int_{0}^{\frac{1}{4}} (4x + 1 - 4x^\frac{1}{2}) dx ) = }$

${\bf \frac{1}{4} * ( (\frac{4}{3} x^3 + 2 x^2 + x)|_{-\frac{1}{2}}^{0} + }$ ${\bf (2 x^2 + x - \frac{8}{3} x^\frac{3}{2} )|_{0}^{\frac{1}{4}} ) = \bf \frac{5}{96} = \frac{m}{n} }$ and ${\bf m + n = 101.}$