Given the functions g(x)=(sqrt2x-1) Use the furmal definition of a derivative at a point to find g'(5). Write the equation of the tangent line (in point slope form) of g(x) at x=5

To solve this problem, we'll proceed in two steps:

1. **Find \( g'(5) \)** using the formal definition of the derivative at a point.
2. **Write the equation of the tangent line** at \( x = 5 \) in point-slope form.

---

### Step 1: Find \( g'(5) \) using the formal definition of the derivative

The function given is:

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

The formal definition of the derivative at a point \( x = a \) is:

\[
g'(a) = \lim_{h \to 0} \frac{g(a + h) - g(a)}{h}
\]

In our case, \( a = 5 \). So, we'll compute \( g'(5) \).

#### Step 1.1: Compute \( g(5) \)

\[
g(5) = \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3
\]

#### Step 1.2: Apply the definition of the derivative

Now we use the definition of the derivative:

\[
g'(5) = \lim_{h \to 0} \frac{g(5 + h) - g(5)}{h}
\]

Substitute \( g(x) = \sqrt{2x - 1} \):

\[
g'(5) = \lim_{h \to 0} \frac{\sqrt{2(5 + h) - 1} - 3}{h}
\]
\[
g'(5) = \lim_{h \to 0} \frac{\sqrt{10 + 2h - 1} - 3}{h}
\]
\[
g'(5) = \lim_{h \to 0} \frac{\sqrt{9 + 2h} - 3}{h}
\]

To evaluate this limit, we rationalize the numerator by multiplying the top and bottom by the conjugate:

\[
g'(5) = \lim_{h \to 0} \frac{\left( \sqrt{9 + 2h} - 3 \right)}{h} \times \frac{\left( \sqrt{9 + 2h} + 3 \right)}{\left( \sqrt{9 + 2h} + 3 \right)}
\]
\[
g'(5) = \lim_{h \to 0} \frac{(9 + 2h) - 9}{h \left( \sqrt{9 + 2h} + 3 \right)}
\]
\[
g'(5) = \lim_{h \to 0} \frac{2h}{h \left( \sqrt{9 + 2h} + 3 \right)}
\]

Simplify:

\[
g'(5) = \lim_{h \to 0} \frac{2}{\sqrt{9 + 2h} + 3}
\]

As \( h \to 0 \), \( \sqrt{9 + 2h} \to \sqrt{9} = 3 \). Therefore, the limit becomes:

\[
g'(5) = \frac{2}{3 + 3} = \frac{2}{6} = \frac{1}{3}
\]

So, \( g'(5) = \frac{1}{3} \).

---

### Step 2: Write the equation of the tangent line

The equation of the tangent line at \( x = 5 \) is given by the point-slope form:

\[
y - y_1 = m(x - x_1)
\]

Where:
- \( m = g'(5) = \frac{1}{3} \) is the slope of the tangent line,
- \( (x_1, y_1) = (5, g(5)) = (5, 3) \) is the point of tangency.

Substitute these values into the point-slope form:

\[
y - 3 = \frac{1}{3}(x - 5)
\]

Thus, the equation of the tangent line is:

\[
y - 3 = \frac{1}{3}(x - 5)
\]

---

Let me know if you'd like more details or if you have any questions.

---

### 5 Related Questions:
1. How is the formal definition of a derivative applied to different functions?
2. How would the process change if \( g(x) \) had a more complex form?
3. Can you find \( g'(x) \) for a general \( x \) in this function?
4. What is the significance of the tangent line at a particular point?
5. How do we determine the limit if the numerator involves more complicated expressions?

### Tip:
When rationalizing expressions involving square roots, multiplying by the conjugate can simplify limits and help you evaluate derivatives.

Given the functions g(x)=(sqrt(2x-1)) Use the formal definition of a derivative at a point to find g'(5). Write the equation of the tangent line (in point slope form) of g(x) at x=5.

Learn how to find the derivative of the function g(x) = sqrt(2x - 1) at x = 5 using the formal definition of a derivative. This detailed solution also shows how to write the equation of the tangent line at this point. Perfect for students in Grades 11-12 or Calculus 1.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation