91Ó°ÊÓ

In calculus, when you need to find the derivative of a fraction where both the numerator and the denominator are differentiable functions, you use the quotient rule. It's handy for dealing with rational functions like our given function: \[y = \frac{2x^2 - x + 3}{x^2 + 2}\] The quotient rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their quotient is: \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\] In this formula, \(u'\) is the derivative of \(u(x)\) and \(v'\) is the derivative of \(v(x)\). For our function, \(u(x) = 2x^2 - x + 3\) and \(v(x) = x^2 + 2\). So, using the quotient rule, the derivative is calculated as follows: \[\frac{dy}{dx} = \frac{(4x - 1)(x^2 + 2) - (2x^2 - x + 3)(2x)}{(x^2 + 2)^2}\] This step might look complex, but the rule simplifies solving derivative questions, especially when functions are divided.

The domain of a function is the set of all possible inputs (x-values) for which the function is defined. For functions involving fractions, checking the denominator is key because dividing by zero isn’t allowed. In our function: \[y = \frac{2x^2 - x + 3}{x^2 + 2}\] The denominator \(x^2 + 2\) doesn’t produce zero for any real number, ensuring the function is defined everywhere. - Since \(x^2 + 2 > 0\) for all real numbers, there are no restrictions on x. - Therefore, the domain is all real numbers: \((-\infty, \infty)\). Understanding the domain helps ensure you consider all possible values in problems of calculus.

End behavior describes how a function behaves as its inputs either become very large or very small. For our function, considering end behavior involves examining the values as \(x\to\infty\) and \(x\to-\infty\). For instance, as we evaluate:\[\lim_{x \to \pm \infty} \frac{2x^2 - x + 3}{x^2 + 2}\] We simplify by dividing everything in the numerator and denominator by \(x^2\): \[\lim_{x \to \pm \infty} \frac{2 - \frac{1}{x} + \frac{3}{x^2}}{1 + \frac{2}{x^2}}\] As \(x\) becomes very large, the terms \(\frac{1}{x}\) and \(\frac{3}{x^2}\) approach zero, leaving us with \(\frac{2}{1} = 2\). Thus, the function approaches the value 2, regardless of whether \(x\) becomes positive or negative infinite.

Understanding limits at infinity is crucial for understanding the end behavior of a function. This concept involves determining what value a function approaches as \(x\) gets extremely large or small. For our function: \[y = \frac{2x^2 - x + 3}{x^2 + 2}\] As \(x\) approaches plus or minus infinity, the terms with smaller powers of \(x\) (like \(x\) or constants) become negligible compared to the highest power. We analyze: \[\lim_{x \to \pm \infty} \frac{2x^2 - x + 3}{x^2 + 2}\] Simplifying by the highest degree of x gives: \[\lim_{x \to \pm \infty} \frac{2 - \frac{1}{x} + \frac{3}{x^2}}{1 + \frac{2}{x^2}} = 2\] This calculation shows that the function's value gets closer and closer to 2, which is the horizontal asymptote.