The domain of the function f(x) = frac{cos^{- | vedclass

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(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The argument of $\cos^{-1}$ must be in $[-1, 1]$: $-1 \leq \

Vedclass · Accepted Answer

$[-\frac{1}{2}, 1) \cup (2, \infty) - \{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\}$

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(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The argument of $\cos^{-1}$ must be in $[-1, 1]$: $-1 \leq \frac{x^{2}-5x+6}{x^{2}-9} \leq 1$.Solving $\frac{x^{2}-5x+6}{x^{2}-9} - 1 \leq 0 \implies \frac{(x-2)(x-3)}{(x-3)(x+3)} - 1 \leq 0 \implies \frac{x-2}{x+3} - 1 \leq 0 \implies \frac{-5}{x+3} \leq 0 \implies x+3 > 0 \implies x > -3$.Solving $\frac{x^{2}-5x+6}{x^{2}-9} + 1 \geq 0 \implies \frac{(x-2)(x-3)}{(x-3)(x+3)} + 1 \geq 0 \implies \frac{x-2}{x+3} + 1 \geq 0 \implies \frac{2x+1}{x+3} \geq 0 \implies x \in (-\infty, -3) \cup [-\frac{1}{2}, \infty)$.Taking the intersection,we get $x \in [-\frac{1}{2}, \infty)$ (excluding $x=3$ where the expression is undefined).$2$. The argument of $\log_{e}$ must be positive: $x^{2}-3x+2 > 0 \implies (x-1)(x-2) > 0 \implies x \in (-\infty, 1) \cup (2, \infty)$.$3$. The denominator cannot be zero: $\log_{e}(x^{2}-3x+2) 
eq 0 \implies x^{2}-3x+2 
eq 1 \implies x^{2}-3x+1 
eq 0 \implies x 
eq \frac{3 \pm \sqrt{5}}{2}$.Combining all conditions: $x \in [-\frac{1}{2}, 1) \cup (2, \infty) - \{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\}$.

The domain of the function $f(x) = \frac{\cos^{-1}\left(\frac{x^{2}-5x+6}{x^{2}-9}\right)}{\log_{e}(x^{2}-3x+2)}$ is:

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