https://file.aiquickdraw.com/mathbot/file/1727049308987sy2xtadu.jpg

We are tasked with finding the equation of the tangent line to the function

\[
f(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}
\]

at \(x = 4\). The equation of the tangent line can be written in the form:

\[
y - f(a) = f'(a)(x - a)
\]

where \(a = 4\), \(f(a)\) is the function value at \(x = 4\), and \(f'(a)\) is the derivative of the function at \(x = 4\).

### Step 1: Compute \(f(4)\)

Substitute \(x = 4\) into the function:

\[
f(4) = \frac{\sqrt{4} + 1}{\sqrt{4} - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3
\]

### Step 2: Compute the derivative \(f'(x)\)

We will use the quotient rule to differentiate \(f(x)\), which states that for \(f(x) = \frac{g(x)}{h(x)}\):

\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}
\]

Here, \(g(x) = \sqrt{x} + 1\) and \(h(x) = \sqrt{x} - 1\).

#### Step 2.1: Derivatives of \(g(x)\) and \(h(x)\)

The derivative of \(g(x) = \sqrt{x} + 1\) is:

\[
g'(x) = \frac{1}{2\sqrt{x}}
\]

The derivative of \(h(x) = \sqrt{x} - 1\) is:

\[
h'(x) = \frac{1}{2\sqrt{x}}
\]

#### Step 2.2: Apply the quotient rule

Now apply the quotient rule:

\[
f'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(\sqrt{x} - 1) - (\sqrt{x} + 1)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x} - 1)^2}
\]

We can simplify the numerator:

\[
f'(x) = \frac{\frac{1}{2\sqrt{x}} \left( (\sqrt{x} - 1) - (\sqrt{x} + 1) \right)}{(\sqrt{x} - 1)^2}
\]

The terms inside the parentheses simplify to:

\[
(\sqrt{x} - 1) - (\sqrt{x} + 1) = -2
\]

So the derivative becomes:

\[
f'(x) = \frac{-\frac{1}{\sqrt{x}}}{(\sqrt{x} - 1)^2}
\]

### Step 3: Evaluate \(f'(4)\)

Substitute \(x = 4\) into \(f'(x)\):

\[
f'(4) = \frac{-\frac{1}{\sqrt{4}}}{(\sqrt{4} - 1)^2} = \frac{-\frac{1}{2}}{(2 - 1)^2} = \frac{-\frac{1}{2}}{1^2} = -\frac{1}{2}
\]

### Step 4: Equation of the tangent line

Now we can write the equation of the tangent line at \(x = 4\):

\[
y - 3 = -\frac{1}{2}(x - 4)
\]

Simplifying:

\[
y - 3 = -\frac{1}{2}x + 2
\]

\[
y = -\frac{1}{2}x + 5
\]

Thus, the equation of the tangent line is:

\[
y = -\frac{1}{2}x + 5
\]

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What is the geometric meaning of a tangent line?
2. How does the quotient rule differ from the product rule in calculus?
3. Can the tangent line equation be found without using derivatives?
4. How would the equation change if \(f(x)\) was a different rational function?
5. Why is it important to understand the slope of a tangent line in real-world applications?

**Tip:** Always remember to check the domain of the function when calculating derivatives, as some points may cause undefined behavior in rational functions.

Write the equation of the tangent line to f(x) = (sqrt(x) + 1) / (sqrt(x) - 1) at x = 4.

Learn how to find the tangent line to the function f(x) = (sqrt(x) + 1) / (sqrt(x) - 1) at x = 4 using the quotient rule for derivatives and the tangent line formula. Suitable for high school and early college students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation