Problem 57 - Calculus and its applications

Question

Find an equation of the tangent line to the graph of \(f(x)=x^{2}- sqrt{x}\). a) at (1,0) ; b) at (4,14) c) at (9,78) .

Step-by-Step Solution

Answer

Tangent lines are: a) \( y = \frac{3}{2}x - \frac{3}{2} \). b) \( y = \frac{31}{4}x - 17 \). c) \( y = \frac{107}{6}x - \frac{155}{2} \).\)

Step 1: Understanding the Problem

We need to find the equation of the tangent line to the function \( f(x) = x^2 - \sqrt{x} \) at three specific points: (1,0), (4,14), and (9,78). The equation of a tangent line at any point \( (x_0, y_0) \) is given by \( y - y_0 = m(x - x_0) \), where \( m \) is the derivative of the function at \( x_0 \), denoted as \( f'(x_0) \).

Step 2: Determine the Derivative

To find the slope of the tangent line, we need the derivative of the function \( f(x) = x^2 - \sqrt{x} \). The derivative is calculated as follows: \( f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(\sqrt{x}) = 2x - \frac{1}{2}x^{-1/2} \).

Step 3: Find the Slopes at Given Points

Compute the derivative at each specified point to find the slopes: - At \( x = 1 \): \( f'(1) = 2 \times 1 - \frac{1}{2} \times 1^{-1/2} = 2 - \frac{1}{2} = \frac{3}{2} \).- At \( x = 4 \): \( f'(4) = 2 \times 4 - \frac{1}{2} \times 4^{-1/2} = 8 - \frac{1}{4} = \frac{31}{4} \).- At \( x = 9 \): \( f'(9) = 2 \times 9 - \frac{1}{2} \times 9^{-1/2} = 18 - \frac{1}{6} = \frac{107}{6} \).

Step 4: Write Tangent Line Equations

Using the formula for the tangent line \( y - y_0 = m(x - x_0) \), substitute the points and their respective slopes:- At (1,0): Slope \( m = \frac{3}{2} \) \[ y - 0 = \frac{3}{2}(x - 1) \Rightarrow y = \frac{3}{2}x - \frac{3}{2} \].- At (4,14): Slope \( m = \frac{31}{4} \) \[ y - 14 = \frac{31}{4}(x - 4) \Rightarrow y = \frac{31}{4}x - 17 \].- At (9,78): Slope \( m = \frac{107}{6} \) \[ y - 78 = \frac{107}{6}(x - 9) \Rightarrow y = \frac{107}{6}x - \frac{155}{2} \].

Key Concepts

Derivative Calculation

To find the tangent line equation, we first need to calculate the derivative of the function. The derivative gives us the slope of the tangent line at any given point on the function. For the function defined as \( f(x) = x^2 - \sqrt{x} \), we need to differentiate both components of the equation separately.

• The derivative of \( x^2 \) is \( 2x \), which is straightforward using the power rule. This rule tells us to bring down the exponent as a coefficient and subtract one from the exponent.
• The derivative of \( \sqrt{x} \), or \( x^{1/2} \), is \( \frac{1}{2}x^{-1/2} \). Applying the power rule again, the exponent \( 1/2 \) comes down as a coefficient, and we subtract one from the exponent.

So, when we compute the derivative of \( f(x) \), it results in: \[ f'(x) = 2x - \frac{1}{2}x^{-1/2} \] This derivative function will be used to find the slope of the tangent at any point \( (x_0, y_0) \). For any point, substitute \( x_0 \) into the derivative to find the slope.

Slope of Tangent Line

The slope of a tangent line is critical, as it determines the direction and steepness of the line at a specific point. Using the derivative \( f'(x) = 2x - \frac{1}{2}x^{-1/2} \), we can find the slope of the tangent line at each provided point. Let’s examine each one together:

• At \( x = 1 \): Substitute into the derivative: \( f'(1) = 2 \times 1 - \frac{1}{2} \times 1^{-1/2} = 2 - \frac{1}{2} = \frac{3}{2} \).
• At \( x = 4 \): Substitute: \( f'(4) = 2 \times 4 - \frac{1}{2} \times 4^{-1/2} = 8 - \frac{1}{4} = \frac{31}{4} \).
• At \( x = 9 \): Substitute: \( f'(9) = 2 \times 9 - \frac{1}{2} \times 9^{-1/2} = 18 - \frac{1}{6} = \frac{107}{6} \).

These calculated slopes define how sharply the function's graph is changing at each of those points. A greater slope magnitude indicates a steeper tangent line.

Equation of Tangent Line

Developing the equation for a tangent line involves using the point-slope form: \[ y - y_0 = m(x - x_0) \] Here, \( m \) is the slope of the tangent line we calculated using the derivative, and \( (x_0, y_0) \) are the coordinates of the point of tangency. Let's write out the equation for each point:

• At (1,0): With slope \( m = \frac{3}{2} \), we substitute into the equation: \[ y - 0 = \frac{3}{2}(x - 1) \] Simplifying, we find: \[ y = \frac{3}{2}x - \frac{3}{2} \]
• At (4,14): With slope \( m = \frac{31}{4} \), the equation becomes: \[ y - 14 = \frac{31}{4}(x - 4) \] Simplifying, we have: \[ y = \frac{31}{4}x - 17 \]
• At (9,78): With slope \( m = \frac{107}{6} \), our equation looks like: \[ y - 78 = \frac{107}{6}(x - 9) \] And simplifying gives us: \[ y = \frac{107}{6}x - \frac{155}{2} \]

This process shows how the slope at each point leads directly to constructing the tangent line's equation, which touches the curve at just that point.