The function f(x) = begin{cases} x - 1, & x

Question

(A) To check the continuity of the function $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2

Vedclass · Accepted Answer

For all real values of $x$

Vedclass · Answer

(A) To check the continuity of the function $f(x)$ at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$. $1$. Left-hand limit: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (x - 1) = 2 - 1 = 1$. $2$. Right-hand limit: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (2x - 3) = 2(2) - 3 = 4 - 3 = 1$. $3$. Value of the function at $x = 2$: $f(2) = 2(2) - 3 = 1$. Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2) = 1$,the function is continuous at $x = 2$. For $x For $x > 2$,$f(x) = 2x - 3$,which is also a polynomial function and thus continuous. Therefore,the function $f(x)$ is continuous for all real values of $x$.

The function $f(x) = \begin{cases} x - 1, & x < 2 \\ 2x - 3, & x \ge 2 \end{cases}$ is a continuous function:

Similar Questions

If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right) & x \neq 0 \\ k & x = 0 \end{cases}$ is continuous,then $k$ is equal to

If $f(x) = \frac{x - e^x + \cos 2x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then which of the following is true? (Note: $[x]$ and $\{x\}$ denote the greatest integer and fractional part functions,respectively.)

If $f(x) = x^3 + 7x - 1$,then $f(x)$ has a zero between $x = 0$ and $x = 1$. The theorem which best describes this is:

Let $f(x) = \frac{1 - \tan x}{4x - \pi }, x \ne \frac{\pi }{4}, x \in [0, \frac{\pi }{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi }{2}]$,then $f(\frac{\pi }{4})$ is

Let $f(x) = \begin{cases} x^3 - x^2 + 10x - 5, & x \le 1 \\ -2x + \log_2(b^2 - 2), & x > 1 \end{cases}$. The set of values of $b$ for which $f(x)$ has the greatest value at $x = 1$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module