well, is the function continuous? yes, is a cubic one, you can graph it if you wish, is continuous all the way, and of course at [ 0, 4] too.
is it differentiable? you can always look at the graph between 0,4 and is a smooth transition line, thus yes, it is differentiable, but let's check anyway,
[tex]\bf \cfrac{dy}{dx}=3x^2-2x-12[/tex] its derivative has no asymptotes and therefore no "cusps", so yes, is differentiable all around.
is f(0) = f(4), let's check
f(0) = 0+0+0+3, f(0) = 3
f(4) = 64 - 16 - 48 + 3, f(4) = 3
yeap
there must then be a "c" value(s) with a horizontal tangent slope, let's check, is really just the critical points.
[tex]\bf \cfrac{dy}{dx}=3x^2-2x-12\implies 0=3x^2-2x-12
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\textit{using the quadratic formula}
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x=\cfrac{-(-2)\pm\sqrt{(-2)^2-4(3)(-12)}}{2(3)}\implies x=\cfrac{2\pm\sqrt{4+144}}{6}
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x=\cfrac{2\pm 2\sqrt{37}}{6}\implies x=\cfrac{2\pm\sqrt{37}}{3}\impliedby \textit{c's}[/tex]
