Question

let ( f(x) = sqrt{4x} ). using the definition of derivative at a point, ( f(a) = limlimits_{x \to a} \frac{f(x) - f(a)}{x - a} ), enter the expression needed to find the derivative at ( x = 3 ). ( f(3) = limlimits_{x \to 3} square ) after evaluating this limit, we see that ( f(3) = square ) finally, the equation of the tangent line to ( f(x) ) where ( x = 3 ) is ( square )

Explanation:

Part 1: Find the expression for \( f'(3) \) using the definition of the derivative

Step 1: Identify \( f(x) \) and \( a \)

We have \( f(x) = \sqrt{4x} \) and we want to find the derivative at \( x = 3 \), so \( a = 3 \). First, find \( f(3) \):
\( f(3) = \sqrt{4 \times 3} = \sqrt{12} = 2\sqrt{3} \)

Step 2: Substitute into the derivative definition

The definition of the derivative at a point \( a \) is \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \). Substituting \( a = 3 \), \( f(x) = \sqrt{4x} \), and \( f(3) = 2\sqrt{3} \) into the formula, we get:
\( f'(3) = \lim_{x \to 3} \frac{\sqrt{4x} - 2\sqrt{3}}{x - 3} \)

Part 2: Evaluate the limit to find \( f'(3) \)

Step 1: Simplify the numerator

First, simplify \( \sqrt{4x} \) to \( 2\sqrt{x} \). So the numerator becomes \( 2\sqrt{x} - 2\sqrt{3} \). We can factor out a 2: \( 2(\sqrt{x} - \sqrt{3}) \).

Step 2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x} + \sqrt{3} \):
\[

$$\begin{align*} \lim_{x \to 3} \frac{2(\sqrt{x} - \sqrt{3})}{x - 3} \times \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} &= \lim_{x \to 3} \frac{2((\sqrt{x})^2 - (\sqrt{3})^2)}{(x - 3)(\sqrt{x} + \sqrt{3})} \\ &= \lim_{x \to 3} \frac{2(x - 3)}{(x - 3)(\sqrt{x} + \sqrt{3})} \end{align*}$$

\]

Step 3: Cancel out \( x - 3 \)

Since \( x \to 3 \) but \( x
eq 3 \), we can cancel the \( x - 3 \) terms:
\( \lim_{x \to 3} \frac{2}{\sqrt{x} + \sqrt{3}} \)

Step 4: Substitute \( x = 3 \)

Now substitute \( x = 3 \) into the expression:
\( \frac{2}{\sqrt{3} + \sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \) (rationalizing the denominator)

Part 3: Find the equation of the tangent line at \( x = 3 \)

Step 1: Find the point of tangency

First, find \( f(3) \): \( f(3) = \sqrt{4 \times 3} = 2\sqrt{3} \), so the point is \( (3, 2\sqrt{3}) \).

Step 2: Use the point - slope form of a line

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point of tangency and \( m \) is the slope (which is \( f'(3) \)). We know \( x_1 = 3 \), \( y_1 = 2\sqrt{3} \), and \( m=\frac{\sqrt{3}}{3} \).
Substitute these values into the point - slope form:
\( y - 2\sqrt{3}=\frac{\sqrt{3}}{3}(x - 3) \)

Step 3: Simplify the equation

Distribute the \( \frac{\sqrt{3}}{3} \) on the right - hand side:
\( y - 2\sqrt{3}=\frac{\sqrt{3}}{3}x-\sqrt{3} \)
Add \( 2\sqrt{3} \) to both sides:
\( y=\frac{\sqrt{3}}{3}x-\sqrt{3}+2\sqrt{3}=\frac{\sqrt{3}}{3}x+\sqrt{3} \)

Answer:

s:

• The expression for \( f'(3) \) is \( \boldsymbol{\frac{\sqrt{4x}-2\sqrt{3}}{x - 3}} \)
• \( f'(3)=\boldsymbol{\frac{\sqrt{3}}{3}} \)
• The equation of the tangent line is \( \boldsymbol{y=\frac{\sqrt{3}}{3}x+\sqrt{3}} \) (or equivalent forms, e.g., \( y=\frac{1}{\sqrt{3}}x+\sqrt{3} \))