Question

a. For the following functions, find \(f^{\prime}\) using the definition. b. Determine an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\) $$f(x)=\sqrt{3 x+1} ; a=8$$

Short answer

Question: Find the equation of the tangent line to the graph of the function \(f(x) = \sqrt{3x + 1}\) at \(x = 8\). Answer: The equation of the tangent line to the graph of the function at \(x = 8\) is: \(y = \frac{3}{10}(x - 8) + 5\).

Step-by-step solution

Find the first derivative of the function using the definition of derivative

The definition of the derivative is: $$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ For our function: $$f(x) = \sqrt{3 x + 1}$$ We will first plug in \((x + h)\) into our function: $$f(x+h) = \sqrt{3(x + h) + 1}$$ Now, we will apply the definition of the derivative: $$f^{\prime}(x) = \lim_{h \to 0} \frac{\sqrt{3(x + h) + 1} - \sqrt{3x+1}}{h}$$

Calculate the first derivative of the function

To find the derivative, we'll first multiply the numerator and denominator by the conjugate of the numerator - this will help to eliminate the square root: $$f^{\prime}(x) = \lim_{h \to 0} \frac{\sqrt{3(x + h) + 1} - \sqrt{3x+1}}{h} \times \frac{\sqrt{3(x + h) + 1} + \sqrt{3x+1}}{\sqrt{3(x + h) + 1} + \sqrt{3x+1}}$$ This simplifies to: $$f^{\prime}(x) = \lim_{h \to 0} \frac{3(x + h) + 1 - 3x - 1}{h(\sqrt{3(x + h) + 1} + \sqrt{3x+1})}$$ After simplifying further, we get: $$f^{\prime}(x) = \lim_{h \to 0} \frac{3h}{h(\sqrt{3(x + h) + 1} + \sqrt{3x+1})}$$ With the common factor h in the numerator and denominator, we can cancel it out. $$f^{\prime}(x) = \lim_{h \to 0} \frac{3}{\sqrt{3(x + h) + 1} + \sqrt{3x+1}}$$ Plugging in the limit, \(h \to 0\), we get: $$f^{\prime}(x) = \frac{3}{\sqrt{3x + 1} + \sqrt{3x+1}}$$ Finally, we get the first derivative of the function: $$f^{\prime}(x) = \frac{3}{2\sqrt{3x + 1}}$$

Determine the tangent line's equation

Now we can find the coordinates of the point where the tangent line touches the graph when \(a=8\). First, plug in \(a=8\) into the function \(f(x)\): $$f(8) = \sqrt{3(8) + 1} = \sqrt{25} = 5$$ Next, find the slope of the tangent line (derivative) at \(x=8\): $$f^{\prime}(8) = \frac{3}{2\sqrt{3(8) + 1}} = \frac{3}{2(5)} = \frac{3}{10}$$ Now, we can use the point-slope form of a linear equation: $$y - y_1 = m(x - x_1)$$ Where \((x_1, y_1) = (8,5)\) is the point on the graph and \(m = \frac{3}{10}\) is the slope of the tangent line. Plug in the values: $$y - 5 = \frac{3}{10}(x - 8)$$ Finally, we get the equation of the tangent line to the graph of the function at \(x=a=8\): $$y = \frac{3}{10}(x - 8) + 5$$

Key concepts

Tangent Line Equation

The tangent line is a straight line that touches a curve at exactly one point. It represents the instantaneous rate of change of the function at that point. To find the equation of a tangent line, we utilize the point-slope form of a line, which is expressed as: \( y - y_1 = m(x - x_1) \). Here,

• \( (x_1, y_1) \) is the specific point where the tangent touches the curve.
• \( m \) is the slope of the tangent line.

To determine \( m \), we use the derivative of the function evaluated at the given point. Once you have \( (x_1, y_1) \) and \( m \), plug them into the equation to find the tangent line.
For example, in the exercise, the point of tangency is \((8, 5)\), and the slope is \(\frac{3}{10}\). Substituting these values gives us:
\( y - 5 = \frac{3}{10}(x - 8) \).
Solving for \(y\), we obtain the equation of the tangent line:
\( y = \frac{3}{10}(x - 8) + 5 \).
This line reflects how the function behaves at that specific spot on the curve.

First Derivative

The first derivative of a function gives us valuable information about the function's rate of change. It's like capturing the steepness or slope of the function at any specific point.

Using the derivative definition, we calculate:

• The formula: \( f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
• This limits process calculates how much \( f(x) \) changes as \( x \) moves a tiny bit.

For the given function \( f(x) = \sqrt{3x + 1} \), applying this definition requires careful algebra:

• Substitute \( x+h \) into \( f(x) \).
• Simplify using conjugates to eliminate the square root.

Finally, cancel common terms to obtain \( f^{\prime}(x) = \frac{3}{2\sqrt{3x + 1}} \).

This derivative helps us understand how fast \( f(x) \) is growing or shrinking at any point \( x \).

Limits in Calculus

Limits form the foundation for understanding derivatives. They tell us about the behavior of functions as values approach a certain point. In calculus, limits allow us to describe instantaneous rates of change.

Consider the limit definition of the derivative:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]Here's what happens:

• \( h \) represents a tiny change in \( x \).
• The fraction \( \frac{f(x+h) - f(x)}{h} \) calculates the average rate of change over the interval \( h \).
• The limit as \( h \to 0 \) gives the precise rate, revealing the function's behavior at that exact point.

By mastering limits, you gain the ability to understand complex functions in a new way. They enable the transition from average change over intervals to understanding specific point behavior, a crucial skill in calculus.