Problem 54

Question

Find the equation of the tangent line to the function \(f\) at the given point. Then graph the function and the tangent line together. $$f(x)=\sqrt{x} \text { at }(4,2)$$

Step-by-Step Solution

Verified
Answer
The equation of the tangent line is \(y = \frac{1}{4}x + 1\).
1Step 1: Find the Derivative
To find the equation of the tangent line, we first need the slope at the given point. Take the derivative of the function \(f(x) = \sqrt{x}\). Using the power rule, the derivative \(f'(x)\) is \(\frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}\).
2Step 2: Evaluate the Derivative at the Point
Substitute \(x = 4\) into the derivative \(f'(x)\) to find the slope of the tangent line at this point: \(f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}\).
3Step 3: Use Point-Slope Form to Find the Equation
The equation of a line can be found using the point-slope formula: \(y - y_1 = m(x - x_1)\). Here, \(m\) is the slope \(\frac{1}{4}\), and the point is \((4, 2)\). Substitute these into the formula: \(y - 2 = \frac{1}{4}(x - 4)\).
4Step 4: Simplify the Equation
Simplify the equation from the previous step to find the equation of the tangent line: \(y - 2 = \frac{1}{4}x - 1\). Therefore, the equation of the tangent line is \(y = \frac{1}{4}x + 1\).
5Step 5: Sketch the Graph
Graph \(f(x) = \sqrt{x}\), which is a curve starting at the origin and increasing steadily as \(x\) increases. Next, plot the tangent line \(y = \frac{1}{4}x + 1\). At \((4, 2)\), both the function and the tangent line should touch, representing where the tangent line approximates the function.

Key Concepts

DerivativePoint-Slope Form
Derivative
The concept of a derivative is crucial to understanding how to find the tangent line to a function at a specific point. A derivative, in simple terms, represents the rate at which a function is changing at any given point. In the context of a curve, it gives us the slope of the tangent line at a point on that curve.

Calculating the derivative involves applying rules like the power rule, product rule, or chain rule, depending on the type of function. For the function given as \(f(x) = \sqrt{x}\), the power rule is applied, resulting in the derivative \(f'(x) = \frac{1}{2\sqrt{x}}\). This derivative shows how steep the curve is, describing how the function behaves as x changes.
  • It helps in determining whether a function is increasing or decreasing.
  • The derivative is essential in solving various optimization problems.
  • In calculus, it's a foundational building block for further studies.
Understanding derivatives empowers you to predict and approximate real-world values accurately.
Point-Slope Form
Once we have the slope from the derivative, we use the point-slope form to find the exact equation of the tangent line. The point-slope form is a straightforward formula used to create a line equation given a point and the slope of that line.

This formula is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope, and \((x_1, y_1)\) represents a specific point on the line.
In our exercise, the slope (\