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

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

Question

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

Vedclass · Accepted Answer

$(-3, -2) \cup (-1, \infty)$

Vedclass · Answer

(D) For the function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ to be defined:$1$. The argument of the logarithm must be positive: $x+3 > 0 \implies x > -3$.$2$. The expression inside the square root in the denominator must be strictly positive: $x^2+3x+2 > 0$.Factoring the quadratic: $(x+2)(x+1) > 0$.This inequality holds when $x \in (-\infty, -2) \cup (-1, \infty)$.Combining the conditions $x > -3$ and $x \in (-\infty, -2) \cup (-1, \infty)$,we get the intersection:$x \in (-3, -2) \cup (-1, \infty)$.

The domain of the real-valued function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ is

Similar Questions

If $f(x) = \tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$ is a real-valued function, then the range of $f$ is:

Find the domain of the function $f(x) = \frac{x^{2}+2x+1}{x^{2}-8x+12}$.

$A$ real valued function $f(x) = |x^2 - 3x + 2| + 2x - 3$ is defined on $[-2, 1]$. If $m$ and $M$ are absolute minimum and absolute maximum values of $f$ respectively, then $M - 4m =$

For $f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$,the domain and range of $f(x)$ in $R$ are:

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module