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

Feb 5, 2018

$k = 1$

#### Explanation:

For $f \left(x\right)$ to be continuous at $x = 3$, the following must be true:

• ${\lim}_{x \to 3} f \left(x\right)$ exists.
• $f \left(3\right)$ 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:

${\lim}_{x \to {3}^{-}} f \left(x\right) = {\lim}_{x \to {3}^{+}} f \left(x\right)$

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

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

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

$\implies {\lim}_{x \to 3} {x}^{3} - 5 {x}^{2} + 2 x + 8 = {\lim}_{x \to 3} - 2 x + k$

Evaluating the limits:

${\left(3\right)}^{3} - 5 {\left(3\right)}^{2} + 2 \left(3\right) + 8 = - 2 \left(3\right) + k$

$\implies 26 - 45 + 6 + 8 = - 6 + k$

$\implies - 5 = - 6 + k$

$\implies k = 1$

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