Problem 17 - Calculus for Biology and Medicine

Question

In Problems \(17-20\), apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\left(3 x^{2}-2\right)(x-1), \text { at } x=1 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Functions U and V

We will apply the product rule, which is used when differentiating products of two functions. Let \( u(x) = 3x^2 - 2 \) and \( v(x) = x - 1 \).

Step 2: Differentiate U and V

Find the derivatives of the two functions: \( u'(x) = \frac{d}{dx}(3x^2 - 2) = 6x \) and \( v'(x) = \frac{d}{dx}(x - 1) = 1 \).

Step 3: Apply the Product Rule

The product rule states that the derivative of \( f(x) = u(x)v(x) \) is \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Applying this: \( f'(x) = (6x)(x - 1) + (3x^2 - 2)(1) \).

Step 4: Simplify the Expression for f'(x)

Simplify the expression: \( f'(x) = 6x(x - 1) + 3x^2 - 2 = 6x^2 - 6x + 3x^2 - 2 \). Combine like terms to get \( f'(x) = 9x^2 - 6x - 2 \).

Step 5: Evaluate f'(x) at x=1

Substitute \( x = 1 \) into \( f'(x) \) to find the slope at \( x = 1 \): \( f'(1) = 9(1)^2 - 6(1) - 2 = 9 - 6 - 2 = 1 \). So, the slope of the tangent line at \( x = 1 \) is 1.

Step 6: Find the Point on the Curve

Substitute \( x = 1 \) into the original function to find the corresponding \( y \)-value: \( f(1) = (3(1)^2 - 2)(1 - 1) = (3 - 2)(0) = 0 \). Therefore, the point on the curve is \((1, 0)\).

Step 7: Write the Equation of the Tangent Line in Slope-Intercept Form

Use the point-slope form of the line: \( y - y_1 = m(x - x_1) \), where \( m = 1 \), \( x_1 = 1 \), \( y_1 = 0 \). Substitute these values: \( y - 0 = 1(x - 1) \). Simplify to get the slope-intercept form: \( y = x - 1 \).

Key Concepts

Product Rule

The product rule is crucial for finding derivatives of functions that are products of two or more simpler functions. When you have a function in the form \( f(x) = u(x) \cdot v(x) \), the product rule helps you determine its derivative \( f'(x) \). The essence of the product rule is simple:

• Differentiate \( u(x) \) to find \( u'(x) \).
• Differentiate \( v(x) \) to find \( v'(x) \).
• Use the formula: \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

In our example, \( u(x) = 3x^2 - 2 \) and \( v(x) = x - 1 \). By applying the rule:

• \( u'(x) = 6x \).
• \( v'(x) = 1 \).
• Thus, \( f'(x) = (6x)(x - 1) + (3x^2 - 2)(1) \).

The product rule allows us to handle more complex functions with ease.

Tangent Line

A tangent line is a straight line that touches a function at a single point, without crossing through it. This line has the same slope as the function at the point of tangency. Finding the tangent line to a curve at a given point involves several steps:

• Determine the derivative of the function to find the slope of the tangent line.
• Evaluate the derivative at the given point to get the exact slope.
• Identify the coordinates of the point on the curve where the tangent intersects.

In the example provided, we found the derivative of the function and evaluated it at \( x = 1 \) to get the slope, which was 1. We then plugged this value into the tangent line equation to solve for the equation of the line.

Slope-Intercept Form

The slope-intercept form of a line is one of the most straightforward methods to write a line equation. It is represented as \( y = mx + b \), where:

• \( m \) is the slope of the line.
• \( b \) is the y-intercept, the point where the line intersects the y-axis.

In constructing the equation of the tangent line:

• Using the slope \( m = 1 \) found earlier, and the point \((1, 0)\) on the curve, we applied the point-slope form to write the equation.
• This involved substituting the values into the equation: \( y - 0 = 1(x - 1) \).
• Solving this, we derived \( y = x - 1 \), which is in slope-intercept form.

The slope-intercept form makes it easy to visualize and sketch the line.

Differentiation Technique

Differentiation is a technique used in calculus to compute the derivative of a function. It measures how a function changes as its input changes. The product rule is one of several differentiation techniques but there are others including:

• The Chain Rule, for differentiating composed functions.
• The Quotient Rule, for functions that are divisions of two other functions.
• Basic differentiation rules like power rules, constants, and sum rules.

In our context, using the product rule is particularly valuable because it allows us to break down and differentiate the product of two functions: \( (3x^2 - 2) \) and \( (x - 1) \). This understanding of differentiation helps in finding slopes, understanding the behavior of functions, and is fundamental to calculus concepts.