Finding a common derivative Two small arches have parabola shapes. The first is described by $f(x)=1-x^{2}$ for $-1<x<1$ and the second is given by $g(x) = 4-(x-4)^{2}$ for $2<x<6$. A board is placed on the arches. Find the slope of the board, given that board does not rest on the tops of either arch.

Use the equality of the expressions for the derivatives $ \ f'(x) \ $ and $ \ g'(x) \ $ to show that $ \ x_1 = x_2 - 4 \ , $ these being the unknown $ \ x-$ coordinates of the tangent points on the left and right parabolas. Write, say, the equation of the tangent line to the left parabola, $ \ y - [1 - x_1^2] \ = \ [f'(x_1)] \ \cdot \ (x - x_1) \ $ ; the tangent point $ \ (x_2 \ , \ 4 - (x_2 - 4)^2 ) \ $ must also be on this line. You should then have the information to find both tangent points; the slope of the board will follow from there.

Here's a picture of the arrangement:

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