Q. 2 - Calculus

Question

Solving for zeroes and non-domain points: For each of the following expressions, find all values of x for which g(x) is zero or does not exist.

$$\begin{gathered} 1. g\left(x\right) = \frac{3x^2-x-2}{x^4+2x^2-3}. \\ 2. g\left(x\right) = \frac{1}{x-2}-\frac{3x+1}{x+2}. \\ 3. g\left(x\right) = \frac{\sin x}{\cos x}. \\ 4. g\left(x\right) = \frac{e^{3x}(x-1)}{\ln x}. \end{gathered}$$

Step-by-Step Solution

Answer

$1. x = -1,-\frac{2}{3}.$

$2. x = 2,-2,-\frac{1}{3}.$

$3. x = n\mathrm{\pi }, \left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2} \mathrm{where} \mathrm{n}\in \mathrm{I}.$

$4. x\in (-\infty ,0].$

Step 1. Given Information.

Given the following functions:

$$\begin{gathered} 1. g\left(x\right) = \frac{3x^2-x-2}{x^4+2x^2-3}. \\ 2. g\left(x\right) = \frac{1}{x-2}-\frac{3x+1}{x+2}. \\ 3. g\left(x\right) = \frac{\sin x}{\cos x}. \\ 4. g\left(x\right) = \frac{e^{3x}(x-1)}{\ln x}. \end{gathered}$$

Step 2. Solving the function for numerator and denominator part (a).

$$\begin{gathered} g\left(x\right) = 0 when numerator is zero i.e., \\ 3x^2-x-2 = 0 \\ x = 1,\frac{-2}{3}. \\ and g\left(x\right)\hspace{0.17em}does not exist when denominator is zero so, \\ {x}^4+2x^2-3 = 0 \\ {x}^2 = 1,-3. \\ Since, square cannot be negative so \\ {x}^2 = 1, \\ and x = 1,-1. \\ Combining all the values of x we get x = 1,-1,-\frac{2}{3}. \\ Since x=1 is a common root therefore it will be cancelled and we get, \\ x = -1,-\frac{2}{3}. \end{gathered}$$

Step 3. Solving the function for numerator and denominator part (b).

$$\begin{gathered} g\left(x\right) = 0 when numerator is zero i.e., \\ 1 cannot be equal to zero. \\ Now, 3x+1=0 \\ gives x = -\frac{1}{3}. \\ and g\left(x\right)\hspace{0.17em}does not exist when denominator is zero so, \\ x-2=0 or x+2=0 \\ we get, \\ x = 2,-2. \\ Combining all the values of x we get x = 2,-2,-\frac{1}{3}. \end{gathered}$$

Step 4. Solving the function for numerator and denominator part (c).

$$\begin{gathered} g\left(x\right) = 0 when numerator is zero i.e., \\ \sin x = 0 \\ x = n\mathrm{\pi }, \mathrm{where} \mathrm{n}\in \mathrm{I}. \\ and g\left(x\right)\hspace{0.17em}does not exist when denominator is zero so, \\ \cos x = 0 \\ x = \left(2n+1\right)\frac{\mathrm{\pi }}{2}, n\in I. \\ Combining all the values of x we get x = n\mathrm{\pi }, \left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2} \mathrm{where} \mathrm{n}\in \mathrm{I}. \end{gathered}$$

Step 5. Solving the function for numerator and denominator part (d).

$$\begin{gathered} g\left(x\right) = 0 when numerator is zero i.e., \\ {e}^{3x}(x-1) = 0 \\ Since, e^{3x} cannot be zero so, x-1=0 \\ And x=1. \\ and g\left(x\right)\hspace{0.17em}does not exist when denominator is zero so, \\ \ln x = 0. \\ x=1. \\ Combining all the values of x we get x = 1. \\ Since x=1 is a common root therefore it will be cancelled and we get, \\ nothing from here but if we see out of the domain all the negative values of x \\ and x=0 will cause the function g\left(x\right) not exist. \\ Therefore, x\in (-\infty ,0]. \end{gathered}$$