Problem 15 - Calculus Early Transcendentals

Question

Equations of tangent lines by definition (2) a. Use definition (2) ( \(p .\) 129) to find the slope of the line tangent to the graph of \(f\) at \(P\). b. Determine an equation of the tangent line at \(P\). $$f(x)=2 x+1 ; P(0,1)$$

Step-by-Step Solution

Answer

Answer: The slope of the tangent line to the graph of the function at point P is 2, and the equation of the tangent line is y = 2x + 1.

Step 1: Find the derivative of the given function

To find the slope, we need to find the derivative of the given function: $$f(x)=2x+1$$ The derivative of the function is given by $$f'(x)=\frac{d(2x+1)}{dx}$$ Since the derivative of 2x with respect to x is 2 and the derivative of a constant is 0, we get $$f'(x)=2$$

Step 2: Determine the slope of the tangent line at point P

Now we know the derivative of the function, we can find the slope at the tangent line at point P. Since f'(x) is constant (2) for any x, the slope of the tangent line at point P is: $$m=f'(0)=2$$ #b. Determine an equation of the tangent line at P#

Step 3: Use the point-slope form of the equation of a line

The point-slope form of the equation of a line is given by: $$y - y_1 = m (x - x_1)$$ Where (x_1, y_1) is the given point (0, 1) and m is the slope we found in the previous step. Now we can plug in the values to find the equation of the tangent line: $$y - 1 = 2 (x - 0)$$

Step 4: Simplify the equation

We can simplify the equation of the tangent line as follows: $$y - 1 = 2x$$ So, the equation of the tangent line at point P is: $$y = 2x + 1$$

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on studying change. It provides the mathematical framework for understanding the way quantities evolve. In essence, calculus is the tool we use to describe and compute the rates at which things change.

• There are two main branches: differential calculus and integral calculus. Differential calculus deals with the concept of the derivative, which allows us to find rates of change and slopes of curves.
• The beauty of calculus is that it formalizes the way we can approach problems involving motion, growth, area, and much more.

In this exercise, we are interested in differential calculus. By finding the derivative of a given function, we can determine the slope of a tangent line at a specific point.

Derivative

A derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point, effectively giving us the slope of the tangent line to the curve.

• The derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
• To find the derivative of a simple linear function like \( f(x) = 2x + 1 \), we take the derivative of each term separately.
• For a linear function \( ax + b \), the derivative is just the coefficient \( a \) of the \( x \) term, since the derivative of a constant is zero.

Therefore, in our problem, the derivative of \( f(x) = 2x + 1 \) is simply \( f'(x) = 2 \). This tells us that the slope of the tangent line at any point on this line is always 2, demonstrating the consistency of linear functions.

Equation of a Line

The equation of a line conveys how \( y \) changes with \( x \). For linear equations, this is generally in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

• A specific form used for finding the equation of a line when a point and slope are known is the point-slope form: \( y - y_1 = m (x - x_1) \).
• Here, \( (x_1, y_1) \) is any point on the line and \( m \) is the slope. This form is especially useful for writing equations of tangent lines.

In our exercise, we use the point \( P(0,1) \) and the slope \( m = 2 \) to write the equation: \[ y - 1 = 2 (x - 0) \]Simplifying this gives us \( y = 2x + 1 \), which is the equation of the tangent line at point \( P \), matching the original function's equation, confirming the line simply touches the graph at that point without crossing it.