Math Snap

STEP 1

Assumptions1. The function $f(x)$ is given by$$f(x)=\left\{\begin{array}{ll}\frac{\sqrt{x+3}-\sqrt{3 x-1}}{x-} & \text { for } x \neq \\ k & \text { for } x=\end{array}\right.$$ . We need to find the value of $k$ such that $f(x)$ is continuous everywhere.
3. A function is continuous at a point if the limit as $x$ approaches that point from the left is equal to the limit as $x$ approaches that point from the right, and both of these are equal to the function's value at that point.

STEP 2

We need to find the limit of the function as $x$ approaches2. This will give us the value of $k$ that makes the function continuous at $x=2$.
$$\lim{{x \to2}} \frac{\sqrt{x+}-\sqrt{x-1}}{x-2}$$

STEP 3

The above limit is an indeterminate form of type0/0. We can apply L'Hopital's rule, which states that the limit of a quotient of two functions as $x$ approaches a point is equal to the limit of the quotients of their derivatives.
$$\lim{{x \to2}} \frac{\sqrt{x+3}-\sqrt{3x-1}}{x-2} = \lim{{x \to2}} \frac{\frac{d}{dx}(\sqrt{x+3}-\sqrt{3x-1})}{\frac{d}{dx}(x-2)}$$

STEP 4

Compute the derivatives of the numerator and the denominator.
The derivative of the numerator is$$\frac{d}{dx}(\sqrt{x+3}-\sqrt{3x-1}) = \frac{1}{2\sqrt{x+3}} - \frac{3}{2\sqrt{3x-1}}$$The derivative of the denominator is$$\frac{d}{dx}(x-2) =1$$

STEP 5

Substitute the derivatives back into the limit.
$$\lim{{x \to2}} \frac{\frac{1}{2\sqrt{x+3}} - \frac{3}{2\sqrt{3x-1}}}{1}$$

Evaluate the limit as $x$ approaches2.
$$k = \frac{1}{2\sqrt{2+3}} - \frac{3}{2\sqrt{3*2-1}} = \frac{1}{2\sqrt{5}} - \frac{3}{2\sqrt{5}} = -\frac{2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}$$So, the value of $k$ that makes the function continuous at $x=2$ is $-\frac{1}{\sqrt{5}}$.