can someone help me with how to find a number that makes a and b or k so that f is continuous at every point. This is just a general question. No specifc problem.

Question

anonymous · Answer

take the limit, replace $k$ by that number

anonymous · Answer

the question is a little to general to answer precisely, and example would be much more enlightening

anonymous · Answer

damn that was wrong, i meant set 
$$k\times 4+2=16$$ get 
$$k=\frac{7}{2}$$

anonymous · Answer

oh okay i see. thank you so much

anonymous · Answer

let me start again 
$$[f(x) = \left\{
     \begin{array}{lr}
       x^2 & : x \leq 4\
       kx+2 & : x >4
     \end{array}
   \right.$$
for the first one you get $4^2=16$ for the second one you get 
$$k\times 4+2$$ so set 
$$4k+2=16$$ solve for $k$ and get $k=\frac{7}{2}$

anonymous · Answer

so you plug in the top and use that to get the bottom. but it would vary with the different problems

anonymous · Answer

post a different problem and we can solve it, the question is a bit vague

anonymous · Answer

okay if I have f(x)= x^2 if x is less than or equal to 4 and x+k if x is greater than 4
So, for the top I would again get 4^2=16 
then, 4+k=16
solve for k and get k=4

anonymous · Answer

if i solve $4+k=16$ i get $k=12$

anonymous · Answer

but that is the right idea, yes

anonymous · Answer

sorry yeah i divided by mistake

anonymous · Answer

oh okay thanks so much. I really appriciate it.

anonymous · Answer

yw