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