Problem 50 - Applied Calculus

Question

(a) Find the equation of the tangent line to \(f(x)=x^{3}\) at the point where \(x=2\). (b) Graph the tangent line and the function on the same axes. If the tangent line is used to estimate values of the function, will the estimates be overestimates or underestimates?

Step-by-Step Solution

Answer

The tangent line equation is \( y = 12x - 16 \). Estimates will be underestimates.

Step 1: Find the Derivative of the Function

To find the equation of the tangent line, we first need to find the derivative of the function, as the derivative gives the slope of the tangent line. The function given is \( f(x) = x^3 \). The derivative of \( f(x) \), denoted as \( f'(x) \), is found by applying the power rule: \( f'(x) = \frac{d}{dx}[x^3] = 3x^2 \).

Step 2: Evaluate the Derivative at x=2

Next, we evaluate the derivative at the given point \( x = 2 \) to find the slope of the tangent line. Substitute \( x = 2 \) into \( f'(x) \): \( f'(2) = 3(2)^2 = 12 \). Thus, the slope of the tangent line at \( x=2 \) is 12.

Step 3: Find the Function Value at x=2

To find a specific point on the function where \( x = 2 \), substitute \( x = 2 \) into the original function \( f(x) \): \( f(2) = (2)^3 = 8 \). So, the point on the function is \( (2, 8) \).

Step 4: Write the Equation of the Tangent Line

With the slope and a point on the line, use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. Using the slope \( m = 12 \) and point \((2, 8)\), the equation is: \( y - 8 = 12(x - 2) \). Simplifying gives \( y = 12x - 16 \), the equation of the tangent line.

Step 5: Consideration of Overestimate or Underestimate

The tangent line \( y = 12x - 16 \) is a linear approximation of the function over a small interval around \( x = 2 \). Since \( f(x) = x^3 \) is a convex function at \( x=2 \) (as \( f''(x) = 6x > 0 \) for \( x > 0 \)), the tangent line lies below the curve for non-negative \( x \). Therefore, the tangent line provides underestimates for function values.

Key Concepts

Derivative

The derivative is essentially a tool that tells us how a function is changing at any given point. It represents the rate of change or the slope of the function at a certain point and is crucial for finding the equation of a tangent line.

When we have a function, for example,

• \( f(x) = x^3 \)

We find its derivative, denoted as \( f'(x) \), to determine the slope of the tangent line at any point on the function. The process involves rules like the power rule in calculus. The power rule states that if you have \( x^n \), where \( n \) is a constant, the derivative is \( nx^{n-1} \). For our function:

• \( f'(x) = 3x^2 \)

This derivative tells us the slope of the tangent line for any \( x \). At \( x = 2 \), the slope becomes 12, showing how steep the line is climbing at that point.

Slope of a Function

The slope of a function at a particular point gives us an idea of how the function is behaving locally, much like how steep a hill is at a single point when you are hiking.

In the scenario from our exercise, the derivative \( f'(x) = 3x^2 \) was calculated. The next critical step was to find what the slope of the function is exactly at \( x = 2 \). We plugged 2 into \( f'(x) \) and found:

• \( f'(2) = 3(2)^2 = 12 \)

This tells us that at the point where \( x = 2 \), the slope of the function \( f(x) = x^3 \) is 12. That means if you were to draw a tangent line touching the curve at just this point, it would climb upward with a steepness or slope of 12.Using the slope makes it possible to construct the tangent line which tells us about the function behavior in the close vicinity of that point.

Point-Slope Form

The point-slope form of a line is an invaluable formula when you know a point on a line and the slope of that line. It's used to write the equation of a line and requires:

• A known slope \( m \)
• A known point \((x_1, y_1)\)

The formula is:

\[ y - y_1 = m(x - x_1) \] In our tangent line example:

• The slope \( m \) is 12.
• The point \((x_1, y_1)\) is (2, 8), taken from the function value at \( x = 2 \).

Plugging these values into the point-slope formula gives us:
\[ y - 8 = 12(x - 2) \] After simplifying, we find:
\[ y = 12x - 16 \] This is the equation of the tangent line. The point-slope form is handy for quickly building this equation around any function at any particular point.

Linear Approximation

Linear approximation is a technique used to make estimates about the values of a function using the tangent line. Essentially, the tangent line is used as a simple method to approximate the function in a small region around a particular point.

In our exercise case, the tangent line equation \( y = 12x - 16 \) is our linear approximation of \( f(x) = x^3 \) near \( x = 2 \). Since the slope at this point was calculated and used, the tangent line provides a linear estimation nearby.

As it turns out, if \( f(x) = x^3 \), then the curve is convex at \( x = 2 \). For convex functions, tangent lines lie below the curve for \( x > 0 \). Thus, in this situation, any estimates made using the tangent line will be underestimates compared to the actual function values. This illustrates how linear approximation can be a straightforward, yet insightful way to predict function behavior.