Question

Let $f(x)=\left[\begin{array}{ll}\frac{3 x^{2}+2 x-1}{6 x^{2}-5 x+1} & \text { for } x \neq \frac{1}{3} \\ -4 & \text { for } x=\frac{1}{3}\end{array}\right.\( then \)f^{\prime}\left(\frac{1}{3}\right)$ (A) is equal to-9 (B) is equal to \(-27\) (C) is equal to 27 (D) does not exist

Short answer

Based on the given information and the solution process, determine the value of the derivative, \(f'(x)\), at \(x=\frac{1}{3}\). Answer: \(-11\)

Step-by-step solution

Find the derivative of the rational function

First, we need to find the derivative of the rational function \(\frac{3x^2 + 2x - 1}{6x^2 - 5x + 1}\). Using the Quotient Rule, which is given by \(\frac{d}{dx}(\frac{u}{v})=\frac{vu'-uv'}{v^2}\), we can find the derivative as follows: Let \(u(x) = 3x^2 + 2x - 1\) and \(v(x) = 6x^2 - 5x + 1\). Then, find the derivatives \(u'(x)\) and \(v'(x)\): \(u'(x) = 6x + 2\) \(v'(x) = 12x - 5\) Now, apply the Quotient Rule: \(f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v^2(x)} = \frac{(6x^2 - 5x + 1)(6x + 2) - (3x^2 + 2x - 1)(12x - 5)}{(6x^2 - 5x + 1)^2}\)

Find the limit of the derivative as x approaches \(\frac{1}{3}\)

Next, we will find the limit of the derivative \(f'(x)\) as \(x\) approaches \(\frac{1}{3}\). This will tell us whether the derivative at \(x = \frac{1}{3}\) exists or not. Evaluate the limit \(\lim_{x \to \frac{1}{3}}f'(x)\): \(\lim_{x \to \frac{1}{3}}\frac{(6x^2 - 5x + 1)(6x + 2) - (3x^2 + 2x - 1)(12x - 5)}{(6x^2 - 5x + 1)^2}\) Plug in \(x=\frac{1}{3}\) and simplify the resulting expression: $f'(\frac{1}{3}) = \frac{(6(\frac{1}{3})^2 - 5(\frac{1}{3}) + 1)(6(\frac{1}{3}) + 2) - (3(\frac{1}{3})^2 + 2(\frac{1}{3}) - 1)(12(\frac{1}{3}) - 5)}{(6(\frac{1}{3})^2 - 5(\frac{1}{3}) + 1)^2} = \frac{(-3)(4) - (-1)(1)}{1} = -11$ Since the limit of the derivative as \(x\) approaches \(\frac{1}{3}\) exists and is equal to -11, we can conclude that \(f'(\frac{1}{3})\) exists and is equal to -11. This means that none of the given answer choices are correct, as they do not include the value -11.