Question

6. Find the value of $ m $ so that the function \[ f(x)=\left\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x>2 \\ 3 x+m, & x \leq 2 \end{array}\right. \] is continuous at $ x=2 $. 2

          6. Find the value of \( m \) so that the function
\[
f(x)=\left\{\begin{array}{ll}
\frac{x^{2}-9}{x-3}, & x>2 \\
3 x+m, & x \leq 2
\end{array}\right.
\]
is continuous at \( x=2 \).
2

$6. Find the value of m so that the function f(x)={ (x^2-9)/(x-3), x>2 3 x+m, x ≤ 2 . is continuous at x=2. 2$

Added by Meagan M.

Calculus, Early Transcendentals

Dennis G. Zill, Warren S. Wright 4th Edition

Chapter 2

Instant Answer

Solved by Expert Atul Kumar

Step 1

A function $ f(x) $ is continuous at $ x = 2 $ if: \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \] Show more…

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6. Find the value of $ m $ so that the function \[ f(x)=\left\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x>2 \\ 3 x+m, & x \leq 2 \end{array}\right. \] is continuous at $ x=2 $. 2