Question

Is the function given by, f(x)={3x-2 if x≤3 , 10-x if x>3} continuous at x = 3?

Answer 1

Answer:

yes

Step-by-step explanation:

This is a piecewise-defined function because it is defined by two or more equations over a specified domain is. The graph of this function is shown below. So this functions is continuous because its graph is a single unbroken curve. So the function is defined be the line 3x - 2 when x = 3 and the output here is y = 7

Answer 2

Answer:

A. Yes, the function is continuous at x = 3.

Step-by-step explanation:

We are given the function,

$f(x)=\left \{{{3x-2}, x\leq 3 \atop {10-x}, x\geq 3} \right$

We will now find the left side and right side limit of f(x) as $x\rightarrow 3$

So, we have,

Left side limit is $\lim_{x \to 3^{-}} f(x)$.

i.e. $\lim_{x \to 3^{-}} f(x)= \lim_{h \to 0} f(3-h)$

i.e. $\lim_{x \to 3^{-}} f(x)= \lim_{h \to 0} 3(3-h)-2$

i.e. $\lim_{x \to 3^{-}} f(x)= \lim_{h \to 0} 9-3h-2$

i.e. $\lim_{x \to 3^{-}} f(x)= \lim_{h \to 0} 7-3h$

i.e. $\lim_{x \to 3^{-}} f(x)= 7$

Right side limit is $\lim_{x \to 3^{+}} f(x)$.

i.e. $\lim_{x \to 3^{+}} f(x)= \lim_{h \to 0} f(3+h)$

i.e. $\lim_{x \to 3^{+}} f(x)= \lim_{h \to 0} 10-(3+h)$

i.e. $\lim_{x \to 3^{+}} f(x)= \lim_{h \to 0} 10-3-h$

i.e. $\lim_{x \to 3^{+}} f(x)= \lim_{h \to 0} 7-h$

i.e. $\lim_{x \to 3^{+}} f(x)= 7$

Thus, the left side limit and the right side limit are equal.

Hence, the function is continuous at x = 3.