Question

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

Answer 1

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

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.

Answer 2

yes

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