Find the domain of the function f(x) = frac{1 | vedclass

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

Question

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive:$(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6) > 0$I

Vedclass · Accepted Answer

$(-5, -1) \cup (0, 4) \cup (6, \infty)$

Vedclass · Answer

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive:$(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6) > 0$Identify the critical points by setting each factor to zero:$x + 1 = 0 \implies x = -1$$e^x - 1 = 0 \implies x = 0$$x - 4 = 0 \implies x = 4$$x + 5 = 0 \implies x = -5$$x - 6 = 0 \implies x = 6$The critical points are $\{-5, -1, 0, 4, 6\}$.Arrange these points on the number line in increasing order: $-5, -1, 0, 4, 6$.Using the wavy curve method (sign scheme):For $x > 6$,the expression is positive.For $4 For $0 For $-1 For $-5 For $x Since we require the expression to be strictly greater than zero,the domain is the union of intervals where the expression is positive:$x \in (-5, -1) \cup (0, 4) \cup (6, \infty)$.

Find the domain of the function $f(x) = \frac{1}{\sqrt{(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6)}}$.

Similar Questions

$A$ 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 $M - 4m =$

The set of all real values of $x$ for which $f(x)=\sqrt{\frac{|x|-2}{|x|-3}}$ is a well-defined function is

The range of the function $f(x) = \frac{x^2}{x^2+1}$ is

The range of the real valued function $f(x) = \frac{x^2+x+1}{x}$ is

The range of the function $f(x) = (\sin x)^{\sin x}$ defined on $(0, \pi)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module