To find the equation of the tangent line to the function $f(x) = \sqrt{x}$ at $x = 2$, we follow these steps:

Step 1: Compute the derivative of $f(x)$

The derivative of $f(x) = \sqrt{x}$ can be computed using the power rule. Rewriting $\sqrt{x}$ as $x^{1/2}$, we get:

$f'(x) = \frac{d}{dx} \left( x^{1/2} \right) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

Step 2: Find the slope of the tangent line at $x = 2$

The slope of the tangent line is given by the value of the derivative at $x = 2$:

$f'(2) = \frac{1}{2\sqrt{2}} = \frac{1}{2\cdot 1.414} \approx 0.353$

Step 3: Find the point on the curve at $x = 2$

The point on the curve at $x = 2$ is $f(2) = \sqrt{2} \approx 1.414$. So the point is $(2, 1.414)$.

Step 4: Use the point-slope form of the equation of a line

The point-slope form of a line is given by:

$y - y_1 = m(x - x_1)$

Where:

• $m$ is the slope of the tangent line,
• $(x_1, y_1)$ is the point of tangency.

Substitute $m = 0.353$, $x_1 = 2$, and $y_1 = 1.414$:

$y - 1.414 = 0.353(x - 2)$

Step 5: Simplify the equation

Now, distribute and simplify:

$y - 1.414 = 0.353x - 0.706$

$y = 0.353x + 0.708$

Thus, the equation of the tangent line to $f(x) = \sqrt{x}$ at $x = 2$ is approximately:

$y = 0.353x + 0.708$

Summary

The equation of the tangent line is $y = 0.353x + 0.708$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How is the derivative of a square root function generally computed?
2. What is the geometric interpretation of a tangent line?
3. How can the point-slope form of a line be derived?
4. What are other methods to find the equation of a tangent line?
5. Can we use the same process for functions that involve trigonometric or exponential terms?

Tip: The derivative of $\sqrt{x}$ can always be expressed as $\frac{1}{2\sqrt{x}}$, which is helpful for finding slopes of tangent lines for square root functions.