Question

Need help with calculus question quickly!

Answer 1

\[\large\rm |x-3|=\cases{(x-3),\qquad x\ge3\\ \rm -(x-3),\quad x\lt3}\]

Answer 2

So when we're on the `left side of 3`,
Our |x-3| is -(x-3), giving us a function of:\[\large\rm t(x)=2^x-\frac{-(x-3)}{(x-3)}\]Err I guess it would be better to do this with limits.
Approach 3 from the left,
and seperately approach 3 from the right
and see if the two pieces connect.

If they don't, we have a jump discontinuity.

Answer 3

\[\large\rm \lim_{x\to3^-}t(x)=2^{x}-\frac{-(x-3)}{(x-3)}\]Do a cancellation before plugging the 3 in :)

Answer 4

Thanks! @zepdrix

Answer 5

Did you figure out out Princess Jasmine? :o

Answer 6

Yes. I said it was a jump discontinuity. I just needed a starting point.

Answer 7

@zepdrix