Problem 19 - Calculus for Biology and Medicine

Question

Apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=4\left(2 x^{4}+3 x\right)\left(4-2 x^{2}\right), \text { at } x=-1 $$

Step-by-Step Solution

Answer

The tangent line in slope-intercept form is \(y = -216x - 224\).

Step 1: Identify the product rule

The product rule is a formula used to find the derivative of the product of two functions. If you have two functions, \(u(x)\) and \(v(x)\), then the derivative of their product \(uv\) is given by: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). \] Here, we identify \(u(x) = 4(2x^4 + 3x)\) and \(v(x) = (4 - 2x^2)\).

Step 2: Compute the derivatives of the functions

Begin by finding the derivatives of \(u(x)\) and \(v(x)\). For \(u(x) = 4(2x^4 + 3x)\), differentiate to obtain \[ u'(x) = 4(8x^3 + 3) = 32x^3 + 12. \]For \(v(x) = 4 - 2x^2\), the derivative is \[ v'(x) = -4x.\]

Step 3: Apply the product rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the product rule formula to find \(f'(x)\):\[ f'(x) = u'(x)v(x) + u(x)v'(x) = (32x^3 + 12)(4 - 2x^2) + 4(2x^4 + 3x)(-4x). \]

Step 4: Simplify the derivative

Expand and combine like terms to simplify the expression for \(f'(x)\):\[ (32x^3 + 12)(4 - 2x^2) = 128x^3 + 48 - 64x^5 - 24x^2, \]\[ 4(2x^4 + 3x)(-4x) = -32x^5 - 48x^2, \]Combine:\[ f'(x) = (128x^3 + 48 - 64x^5 - 24x^2) + (-32x^5 - 48x^2). \]

Step 5: Evaluate the derivative at the specified point

Substitute \(x = -1\) into \(f'(x)\) to find the slope of the tangent line:First calculate \[ 64(-1)^5 = -64, \ 128(-1)^3 = -128, \ -24(-1)^2 = -24, \ -48(-1)^2 = -48, \ 48(1) = 48. \]Then substitute these into the expression for the derivative \[ f'(-1) = (-128 - 64 - 24 - 48 + 48) = -216. \] This is the slope of the tangent line at \(x = -1\).

Step 6: Find the y-coordinate at the specified x-value

Evaluate \(f(x)\) at \(x = -1\) to find the y-coordinate of the point:\[ f(-1) = 4(2(-1)^4 + 3(-1))(4 - 2(-1)^2) = 4(2 - 3)(4 - 2). \]Simplify to get \[ f(-1) = 4(-1)(2) = -8. \] The point is \((-1, -8)\).

Step 7: Write the tangent line equation in slope-intercept form

Use the point-slope form of the line \(y - y_1 = m(x - x_1)\) and convert it to slope-intercept form \(y = mx + b\):\[ y - (-8) = -216(x - (-1)) \]Simplify to find\[ y + 8 = -216(x + 1) \Rightarrow y = -216x - 216 - 8 \Rightarrow y = -216x - 224. \]

Key Concepts

Calculus

Calculus is a branch of mathematics that explores how things change, particularly by examining rates of change and quantities that accumulate over time. It is primarily divided into two fields:

• Differential Calculus: This focuses on the concept of a derivative, which represents the rate of change of one quantity in relation to another. It is particularly useful in finding slopes of curves and rates of change. In this context, the derivative helps us understand how a function behaves at a specific point.
• Integral Calculus: This deals with the accumulation of quantities, such as areas under curves. It's used to find total quantities given a rate of change.

In the exercise, we used differential calculus to find the slope of the tangent line to a function at a given point. This is accomplished by differentiating the function using the product rule. Understanding calculus is essential for tackling complex mathematical problems and models that describe real-world phenomena.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. At this point, the line has the same slope as the curve. Finding the tangent line is critical for understanding the local behavior of a curve. In practice, it provides an approximation of the curve at the point of contact.

The process to find a tangent line involves:

• Calculating the derivative of the curve, which gives the slope of the tangent line at any given point.
• Identifying the specific point of interest, often expressed as \(x = a\).
• Substituting the point into the derivative to find the slope \(m\) of the tangent line at that point.
• Using the point-slope form of the equation to write the equation of the tangent line.

In the exercise, the tangent line at \(x = -1\) was calculated by first finding the derivative using the product rule and then evaluating the derivative at the specified point to get the slope.

Slope-Intercept Form

The slope-intercept form is a straightforward way of writing the equation of a line. It highlights the slope and y-intercept clearly and is expressed as:\[ y = mx + b \]where \(m\) represents the slope of the line, and \(b\) is the y-intercept, indicating where the line crosses the y-axis.

This form is particularly useful because:

• It provides immediate visual clues about the line's direction and steepness—the slope \(m\) tells us how the line rises over runs between two points.
• It simplifies the process of graphing lines, since only two values are needed.
• It facilitates quick comparison between different lines.

In the exercise, after calculating the slope at \(x = -1\), we used both the slope and a point on the curve to find the line's equation in slope-intercept form. This equation tells us all we need to know about the tangent line's positioning in the Cartesian plane.