Problem 61 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 61-64, find an equation of the tangent line to the graph of the function at the given point. \(f(x)=(1-x)\left(x^{2}-1\right)^{2} ;(2,-9)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of the function \(f(x) = (1-x)(x^2-1)^2\) at the point (2,-9) is \(y = -33x + 57\).

Step 1: Find the derivative of the function

To find the slope of the tangent line, we need to find the function's derivative using the product and chain rules. Function: \(f(x) = (1 - x)(x^2 - 1)^2\) Apply the product rule: \[\frac{d}{dx}\left[u(x)v(x)\right] = u'(x)v(x) + u(x)v'(x)\] Let \(u(x) = 1 - x\) and \(v(x) = (x^2 - 1)^2\) Differentiate \(u(x)\): \[u'(x) = -1\] Differentiate \(v(x)\) using the chain rule: \[\frac{d}{dx}\left[w(x)^2\right] = 2w(x)w'(x)\] Let \(w(x) = x^2 - 1\) Differentiate \(w(x)\): \[w'(x) = 2x\] Now, differentiate \(v(x)\): \[v'(x) = 2(x^2 - 1)(2x) = 4x(x^2 - 1)\] Now, find the derivative using the product rule formula: \[f'(x) = u'(x)v(x) + u(x)v'(x)\] \[f'(x) = (-1)\left((x^2 - 1)^2\right) + (1 - x)\left(4x(x^2 - 1)\right)\]

Step 2: Find the slope of the tangent line at the point (x = 2)

To find the slope of the tangent line at the point (2,-9), substitute x = 2 into the derivative function: \[f'(2) = (-1)\left((2^2 - 1)^2\right) + (1 - 2)\left(4 \cdot 2 (2^2 - 1)\right)\] \[f'(2) = (-1)(3^2) - 4(2)(3)\] \[f'(2) = -9 - 24\] \[f'(2) = -33\] The slope of the tangent line is -33.

Step 3: Write the equation of the tangent line

Now that we have the slope, we can use the point-slope form equation to find the equation of the tangent line: \[y - y_{1} = m(x - x_{1})\] With the point (2, -9) and the slope -33, plug in the values: \[y - (-9) = -33(x - 2)\] Now simplify and solve for y: \[y = -33(x - 2) + (-9)\] \[y = -33x + 66 - 9\] \[y = -33x + 57\] The equation of the tangent line is \(y = -33x + 57\).

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It has the same slope as the curve at that specific point. Finding the tangent line is a common task in calculus because it gives insight into the behavior of the function around the point of interest.
To find the tangent line, we first need the derivative of the function, which tells us the slope of the tangent line. Once the slope is known, we can use the point-slope form of a linear equation to write the equation of the tangent line. This form is expressed as:

• Point-Slope Form: \[ y - y_1 = m(x - x_1) \]
• Where \( (x_1, y_1) \) is the given point on the curve and \( m \) is the slope from the derivative.

In this exercise, we found the slope using the derivative and the point given, leading us to the equation \( y = -33x + 57 \).

Derivative

The derivative of a function measures how the function value changes as its input changes. It represents the slope of the tangent line to the curve at any point. In notation, the derivative of a function \( f(x) \) is often represented as \( f'(x) \) or \( \frac{df}{dx} \).
Calculating the derivative involves using rules and principles such as the power rule, product rule, and chain rule. In this exercise, the derivative is necessary to find the tangent line's slope. The process involves differentiating each part of the function and combining them effectively:

• Use the product rule when dealing with functions multiplied together.
• When a function is a composition of two functions, the chain rule is applied.

For instance, the function \((1-x)(x^2-1)^2\) required careful application of both the product and chain rules to derive \(f'(x)\).

Product Rule

The product rule is a technique used in calculus to differentiate functions that are products of two or more functions. The rule states that the derivative of the product of two functions \( u(x) \) and \( v(x) \) is given by:
\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
This rule is essential in finding the derivative of complex functions, like the one in this exercise. We used the product rule on the function \( (1-x)(x^2-1)^2 \) by letting \( u(x) = 1-x \) and \( v(x) = (x^2-1)^2 \).

• The derivative of \( u(x) \), \( u'(x) = -1 \), was straightforward.
• The derivative of \( v(x) \) required a secondary step involving the chain rule.

Using these, we combined the derivatives to find \( f'(x) \), giving insight into the slope of the tangent line.

Chain Rule

The chain rule is a fundamental rule in calculus for differentiating composite functions. If you have a function \( y = g(f(x)) \), then the chain rule states:
\[ \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \]
In simpler terms, differentiate the outer function while keeping the inner function unchanged, and then multiply by the derivative of the inner function.
In our exercise, \( v(x) = (x^2 - 1)^2 \) required the chain rule. Here, the inner function \( w(x) = x^2 - 1 \) was differentiated first to get \( w'(x) = 2x \).

• The outer function was \( t^2 \), whose derivative is \( 2t \).
• The result of applying the chain rule was: \( 2(x^2 - 1) \cdot 2x = 4x(x^2 - 1) \).

This strategy allowed us to effectively break down and solve the problem of finding \( v'(x) \) within the larger differentiation task. It illustrates how powerful the chain rule is in calculus.