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Continuity

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Find the constant that makes the function continuous

k=2
k=1
k=3
k=(1)/(6)
k=1
k=1
k=frac12
k=12
Removable: lim_xto2f=4 but f(2) does not exist.
Infinite: vertical asymptote at x=3.
Finite jump: one-sided limits -1 and 1.
Essential: it oscillates between -1 and 1 and has no limit.
Finite jump: one-sided limits -1 and 1.
Removable: lim_xto0f=1 but f(0) does not exist.
Yes: f(0)=-1<0<1=f(1) and f is continuous, so by the IVT there is cin(0,1) with f(c)=0.
At every real number except x=-2 and x=3.
a=tfrac13 and b=tfrac23.
Yes: with g(x)=x-cos x, g(0)=-1<0 and g(1)=1-cos1>0; by the IVT a solution exists.
It is continuous since lim_xto2f=0=f(2), but it is not differentiable: the one-sided slopes are -1 and 1.
No: f is not continuous on [-1,1] (it is undefined at x=0), so the IVT does not apply; in fact frac1xneq0 always.