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Find the value of k so that

f(x)= {3x^2+kx+4 if x<0 and 2sqrt(x+4) if x >or equal to 0}

is differentiable at x=0

can someone please provide help!!!

they have the same value at x = 0

so we only need the same slope

on the left
dy/dx = 6 x + k
on the right
dy/dx = 2(1/2) /sqrt(x+4) = 1/sqrt(x+4)
at x = 0
6(0) + k = 1/sqrt(4)
so k = 1/2

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