The domain of f(x) = frac{log_2(x + 3)}{x^2 + | vedclass

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

Question

(D) For the function $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ to be defined:$1$. The argument of the logarithm must be positive: $x + 3 > 0 \implie

Vedclass · Accepted Answer

$(-3, +\infty) - \{-1, -2\}$

Vedclass · Answer

(D) For the function $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ to be defined:$1$. The argument of the logarithm must be positive: $x + 3 > 0 \implies x > -3$.$2$. The denominator must not be zero: $x^2 + 3x + 2 
eq 0$.Factoring the denominator: $(x + 1)(x + 2) 
eq 0$,which implies $x 
eq -1$ and $x 
eq -2$.Combining these conditions,we require $x > -3$ and $x 
otin \{-1, -2\}$.Thus,the domain is $(-3, \infty) - \{-1, -2\}$.

The domain of $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ is

Similar Questions

Domain of the function $f(x) = \sqrt{2 - 2x - x^2}$ is

The domain of the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ is

If $f(x) = a\cos(bx + c) + d$,then the range of $f(x)$ is

Out of $800$ boys in a school,$224$ played cricket,$240$ played hockey,and $336$ played basketball. Of the total,$64$ played both basketball and hockey,$80$ played cricket and basketball,and $40$ played cricket and hockey; $24$ played all three games. The number of boys who did not play any game is:

The shaded region in the given figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module