Question

What is k such that function f is continuous at x = 3?

Answer 1

`k = 1`

Explanation:

For `f(x)` to be continuous at `x = 3`, the following must be true:

• `lim_(x->3)f(x)` exists.
• `f(3)` exists

Let's investigate the first postulate. We know that for a limit to exist, the left hand and right hand limits must be equal. Mathematically:

`lim_(x->3^-)f(x) = lim_(x->3^+)f(x)`

As defined above, we can see that to the left of `x = 3`, `f(x) = x^3 - 5x^2 + 2x + 8`.

Also, to the right of (and including) `x = 3`, `f(x) = -2x + k`

Hence, the limits of these functions as they approach `3` must be equal:

`=> lim_(x->3) x^3 - 5x^2 + 2x + 8 = lim_(x->3) -2x + k`

Evaluating the limits:

`(3)^3 - 5(3)^2 + 2(3) + 8 = -2(3) + k`

`=>26 - 45 + 6 + 8 = -6 + k`

`=> -5 = -6 + k`

`=> k = 1`

`k` being 1 guarantees that the limit exists, and we already know that `f(3)` exists (we actually just evaluated that above), and hence we can say that the function is continuous at `x = 3` IFF `k = 1`.