Problem 17 - Calculus and its applications

Question

Find an equation of the tangent line to the graph of \(f(x)=x^{3}\) at (a) (-2,-8) (b) (0,0) (c) (4,64)

Step-by-Step Solution

Answer

Tangent lines are: (a) \(y = 12x + 16\), (b) \(y = 0\), (c) \(y = 48x - 128\).

Step 1: Find the Derivative of the Function

The first step in finding the equation of the tangent line is to determine the derivative of the function. Here, the function is given as \( f(x) = x^3 \). The derivative of this function, which represents the slope of the tangent, is found using the power rule: \( f'(x) = 3x^2 \).

Step 2: Evaluate the Derivative at the Given Points

For each point, we need to find the slope by evaluating the derivative at the x-value of each point. (a) For \((-2, -8)\), evaluate \(f'(-2) = 3(-2)^2 = 12\).(b) For \((0, 0)\), evaluate \(f'(0) = 3(0)^2 = 0\).(c) For \((4, 64)\), evaluate \(f'(4) = 3(4)^2 = 48\).

Step 3: Use Point-Slope Form to Write the Equation of the Tangent Line

The formula for the equation of a line given a slope \(m\) and a point \((x_1, y_1)\) is the point-slope form: \( y - y_1 = m(x - x_1) \). (a) For \((-2, -8)\), \( m = 12 \): \( y + 8 = 12(x + 2) \) which simplifies to \( y = 12x + 16 \).(b) For \((0, 0)\), \( m = 0 \): \( y - 0 = 0(x - 0) \) which simplifies to \( y = 0 \).(c) For \((4, 64)\), \( m = 48 \): \( y - 64 = 48(x - 4) \) which simplifies to \( y = 48x - 128 \).

Key Concepts

Derivative Calculation

To understand how to find a tangent line to a curve, we first need to calculate the derivative of the function involved. For our problem, the function is given as \( f(x) = x^3 \). The derivative, \( f'(x) \), represents the slope of the curve at any given point. How do we calculate this derivative? By using differentiation rules!

Power Rule

The Power Rule is a fundamental tool for finding derivatives. It states that if you have a function \( f(x) = x^n \), where \( n \) is any constant, then the derivative is \( f'(x) = nx^{n-1} \). It helps simplify complex calculations.
In our example of \( f(x) = x^3 \), we lower the power of \( x \) by one and multiply by the original power: \[ f'(x) = 3x^{3-1} = 3x^2 \]. This derivative \( 3x^2 \) gives us the slope of the tangent line at any point on \( f(x) \).
These rules are not just applicable to this function but a wide range of polynomial expressions.

Point-Slope Form

Once we calculate the slope using the derivative, the next step is to write the equation of the tangent line by using the Point-Slope Form. This form is particularly useful because it allows you to write the equation of a line if you know the slope and any point on the line.

• The Point-Slope Form formula is: \( y - y_1 = m(x - x_1) \).
• Here, \( m \) is the slope, and \((x_1, y_1)\) is a point on the line.

For instance, at the point \((-2, -8)\), with a slope \(m = 12\), the equation becomes \( y + 8 = 12(x + 2) \).
Understanding this form is crucial because it connects slope and points directly, simplifying the process of writing linear equations.

Slope of a Tangent Line

The slope of a tangent line to a curve at a particular point indicates how the curve behaves at that point. The derivative function we derived earlier, \( f'(x) = 3x^2 \), gives us the slope of the tangent line for any value of \( x \).

• The slope tells us whether the function is increasing or decreasing at that point.
• For example: At \((0, 0)\), the slope is \(0\), indicating a horizontal tangent.
• At \((-2, -8)\), the slope is \(12\), indicating a steep upward tangent.
• And at \((4, 64)\), with a slope of \(48\), the tangent is even steeper.

This understanding allows us to predict the direction and steepness of the curve near the points of interest.