The range of the real valued function f(x) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the function $f(x) = |x-2| + |x-3|$.We analyze the function in three intervals: $x \leq 2$, $2 < x < 3$, and $x \geq 3$.For $x \leq 2$, $f(x

Vedclass · Accepted Answer

$[1, \infty)$

Vedclass · Answer

(B) Given the function $f(x) = |x-2| + |x-3|$.We analyze the function in three intervals: $x \leq 2$, $2 For $x \leq 2$, $f(x) = -(x-2) - (x-3) = -x + 2 - x + 3 = -2x + 5$. Since $x \leq 2$, $-2x \geq -4$, so $f(x) \geq -4 + 5 = 1$.For $2 For $x \geq 3$, $f(x) = (x-2) + (x-3) = 2x - 5$. Since $x \geq 3$, $2x \geq 6$, so $f(x) \geq 6 - 5 = 1$.Combining these, the minimum value of the function is $1$ and it takes all values greater than or equal to $1$.Thus, the range is $[1, \infty)$.Therefore, option $(B)$ is correct.

The range of the real valued function $f(x) = |x-2| + |x-3|$ is

Similar Questions

If $[\cdot]$ denotes the greatest integer function,then the domain and range of the function $f(x) = \frac{\sin([x]\pi) + \tan([x]\pi)}{1 + [x]^2 + [x]^4}$ are respectively

The domain of the function $f(x) = \frac{\cos^{-1}\left(\frac{x^{2}-5x+6}{x^{2}-9}\right)}{\log_{e}(x^{2}-3x+2)}$ is:

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x-1}$.

Domain of the real valued function $f(x) = \frac{x+2}{9-x^{2}}$ is

If $[x]^2-5[x]+6=0$,where $[x]$ denotes the greatest integer function,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module