Find the domain of the function f(x) = frac{x | vedclass

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

Question

(A) The function $f(x) = \frac{x^{2}+3x+5}{x^{2}-5x+4}$ is defined for all real values of $x$ except where the denominator is zero.Set the denominator

Vedclass · Accepted Answer

$R - \{1, 4\}$

Vedclass · Answer

(A) The function $f(x) = \frac{x^{2}+3x+5}{x^{2}-5x+4}$ is defined for all real values of $x$ except where the denominator is zero.Set the denominator equal to zero: $x^{2}-5x+4 = 0$.Factor the quadratic expression: $(x-4)(x-1) = 0$.This gives $x = 4$ or $x = 1$.Therefore,the function is undefined at $x = 1$ and $x = 4$.The domain of the function is the set of all real numbers except $1$ and $4$,which is written as $R - \{1, 4\}$.

Find the domain of the function $f(x) = \frac{x^{2}+3x+5}{x^{2}-5x+4}$.

Similar Questions

If $[x]$ denotes the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

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

The sum of the least positive integer and the greatest negative integer in the range of the function $f(x) = \frac{x^2-5x+7}{x^2-5x-7}$ is

The domain of the function $f(x) = \sec^{-1}(3x - 4) + \tanh^{-1}\left(\frac{x + 3}{5}\right)$ is

If the domain of the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ is $(\alpha, \beta) \cup (\gamma, \delta)$,then $\alpha + \beta + \gamma + \delta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module