Problem 42 - Brief Applied Calculus

Question

a. Find the equation of the tangent line to \(f(x)=3 x^{2}-x^{3} \quad\) at \(\quad x=1\) b. Graph the function and the tangent line on the window \([-1,3]\) by \([-2,5]\).

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 3x - 1 \).

Step 1: Find the derivative of the function

To find the equation of the tangent line, we first need the derivative of the function to find the slope. The function is given as \( f(x) = 3x^2 - x^3 \). The derivative, \( f'(x) \), is calculated as follows:\[ f'(x) = \frac{d}{dx}(3x^2 - x^3) = 6x - 3x^2 \]

Step 2: Calculate the slope of the tangent line at \( x = 1 \)

The slope of the tangent line at a given point is the value of the derivative at that point. Substitute \( x = 1 \) into \( f'(x) \) to find the slope:\[ f'(1) = 6(1) - 3(1)^2 = 6 - 3 = 3 \]Thus, the slope of the tangent line at \( x = 1 \) is 3.

Step 3: Determine the point on the curve

The tangent line should pass through the point on the function where \( x = 1 \). Substitute \( x = 1 \) into the original function \( f(x) \) to find the corresponding \( y \)-value:\[ f(1) = 3(1)^2 - (1)^3 = 3 - 1 = 2 \]So, the point is \( (1, 2) \).

Step 4: Write the equation of the tangent line

Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. With the slope \( m = 3 \) and the point \( (1, 2) \), we have:\[ y - 2 = 3(x - 1) \]Simplifying gives:\[ y = 3x - 1 \]This is the equation of the tangent line.

Step 5: Graph the function and tangent line

To graph the function \( f(x) = 3x^2 - x^3 \) and the tangent line \( y = 3x - 1 \), set up your graphing window as specified: \( x \) from \(-1\) to \(3\) and \( y \) from \(-2\) to \(5\). Make sure the graph of the quadratic curve intersects with the linear line at the point \( (1, 2) \), and the line should have a positive slope, matching the calculated slope of 3.

Key Concepts

Derivative

To understand the concept of a tangent line, we first need to understand derivatives. The derivative of a function gives us the rate at which the function's value changes as the input changes. It's like a powerful tool that helps us find how steep a function is at any point. When you take the derivative of a function, mathematically, this is written as \( f'(x) \). It tells you the slope of the tangent to the function at any given point.
In our exercise, the function is \( f(x) = 3x^2 - x^3 \). We found its derivative, which is \( f'(x) = 6x - 3x^2 \). This derivative will help us determine the slope of the tangent line wherever needed.

Slope of a function

The slope of a function at a specific point is particularly important because it tells us how steep the line is at that very moment. Think of it like riding a bicycle uphill or downhill; the steeper the hill, the harder it is to pedal. Similarly, this slope indicates the inclination of the function.
To find the slope of the tangent line at \( x = 1 \) for our function, we substitute \( x = 1 \) into the derivative \( f'(x) \). From the derivative \( 6x - 3x^2 \), substituting \( x = 1 \) gives:

• \( f'(1) = 6(1) - 3(1)^2 = 6 - 3 = 3 \)

This value, 3, represents the slope of the tangent line at the point where \( x = 1 \). It tells us that the line is rising relatively steeply at this point.

Point-slope form

The point-slope form is a straightforward way to write the equation of a line when we know its slope and a point it passes through. This form is expressed in mathematics as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is a known point on the line.
In our scenario, we found the slope \( m \) to be 3, and the point on the curve when \( x = 1 \) is \( (1, 2) \). Plugging these into the point-slope formula gives:

• \( y - 2 = 3(x - 1) \)

Simplifying this equation, we derive:

• \( y = 3x - 1 \)

This equation perfectly describes the tangent line in terms of its slope and a known point. By knowing this form, we can quickly sketch or understand any linear graph passing through a specific point and having a particular slope.