Consider f(x) = 2√(x+2) - 5 f or x ≤ 2
-2/x f or x>2
Does f(x) have a tangent line at x=2? If it does, provide the equation of the line. If it doesn't, clearly explain why. Show work.
f'(x) = 1/√(x+2) for x≤2
f'(x) = 2/x^2 for x>2
lim 2- = 1/2
lim 2+ = 1/2
So far, so good. But
lim 2- f(x) = -4
lim 2+ = -1
Since the limits are not the same, f is not continuous at x=2
So, even though the slopes are the same, there is no single tangent line.