Let f: R rightarrow R be defined as f(x) = fr | vedclass

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(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.Multiplying both sides by $(x^2+x+4)$, we get $y(x^2+x+4) = x^2-x+4$.Rearranging the terms, we get $(y-1)x^2 + (

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$[\frac{3}{5}, \frac{5}{3}]$

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(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.Multiplying both sides by $(x^2+x+4)$, we get $y(x^2+x+4) = x^2-x+4$.Rearranging the terms, we get $(y-1)x^2 + (y+1)x + (4y-4) = 0$.For $x$ to be a real number, the discriminant $D$ must be greater than or equal to $0$.$D = (y+1)^2 - 4(y-1)(4y-4) \geq 0$.$(y+1)^2 - 16(y-1)^2 \geq 0$.Using the identity $a^2 - b^2 = (a-b)(a+b)$, we get $((y+1) - 4(y-1))((y+1) + 4(y-1)) \geq 0$.$(y+1-4y+4)(y+1+4y-4) \geq 0$.$(5-3y)(5y-3) \geq 0$.Multiplying by $-1$ reverses the inequality: $(3y-5)(5y-3) \leq 0$.The roots are $y = \frac{5}{3}$ and $y = \frac{3}{5}$.Thus, the range is $y \in [\frac{3}{5}, \frac{5}{3}]$.

Let $f: R \rightarrow R$ be defined as $f(x) = \frac{x^2-x+4}{x^2+x+4}$. Then, the range of the function $f(x)$ is

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