Problem 7 - Precalculus Mathematics for Calculus

Question

Find the slope of the tangent line to the graph of \(f\) at the given point. $$f(x)=2 x^{3} \quad \text { at }(2,16)$$

Step-by-Step Solution

Answer

The slope of the tangent line at the point \((2,16)\) is 24.

Step 1: Understand the Problem

We need to find the slope of the tangent line to the graph of the function \( f(x) = 2x^3 \) at the point \((2,16)\). To find this slope, we need to compute the derivative of the function, \( f'(x) \), and evaluate it at \( x = 2 \).

Step 2: Differentiate the Function

To find the derivative \( f'(x) \), we use the power rule. The power rule states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \). For \( f(x) = 2x^3 \), this gives us:\[ f'(x) = 3 \cdot 2 \cdot x^{3-1} = 6x^2 \].

Step 3: Evaluate the Derivative at the Given Point

Now that we have the derivative \( f'(x) = 6x^2 \), we evaluate it at \( x = 2 \) to find the slope of the tangent line:\[ f'(2) = 6(2)^2 = 6 \times 4 = 24 \].

Step 4: Conclusion

The slope of the tangent line to the graph of \( f(x) = 2x^3 \) at the point \( (2,16) \) is \( 24 \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus used to measure how a function changes as its input changes. In simpler terms, a derivative represents the slope of a function at any given point on its graph.

• This slope tells you how steep the line is at that point.
• For linear functions, the slope is constant, but for non-linear functions, like curves, the slope varies as you move along the graph.

To find a derivative, we use differentiation, which involves applying specific rules to a given function. In this exercise, we used the derivative to find the slope of the tangent line to the curve of the function at a specific point. Differentiation helps in understanding the rate of change of a function, which is crucial for solving real-world problems such as finding velocities or growth rates. This exercise focuses on computing the derivative of a polynomial function, which is where the power rule comes in handy. Derivatives are not only crucial for understanding slopes and rates of change, but also for optimizing functions, which is useful in economics, physics, and various scientific applications.

Exploring the Power Rule

The power rule is a quick and straightforward way to differentiate polynomial functions. This rule is vital when working with functions of the form \(f(x) = ax^n\). Here's how it works:

• The power rule states that the derivative of \(ax^n\) is \(nax^{n-1}\).
• This means you'll multiply the exponent \(n\) by the coefficient \(a\) and then reduce the exponent by one.

Using the power rule makes differentiating polynomial functions swift and easy. In this exercise, we applied the power rule to the function \(f(x) = 2x^3\). By applying the rule, we found the derivative to be \(f'(x) = 6x^2\). Understanding the power rule is essential as it forms the basis for differentiating more complex functions. If you practice applying the power rule to different polynomial functions, you will soon find it is an indispensable tool in your calculus toolkit. It streamlines the process and reduces the potential for error.

Importance of Function Evaluation

Function evaluation involves substituting specific values into a function to gain useful information, like the value of a function at a particular point or the slope of its tangent.

• Once a derivative is found, evaluating it at a given point allows us to determine the slope of a tangent line.
• This information is crucial in understanding the behavior of the graph at that particular point.

In our exercise, after finding the derivative \(f'(x) = 6x^2\), we evaluate it at \(x = 2\), leading to \(f'(2) = 24\). This tells us the slope of the tangent at the point \((2, 16)\). Function evaluation is not only used for finding tangent slopes, but also for determining rates of change and predicting future outcomes in applied sciences. Moreover, understanding how to properly evaluate functions is a skill that enhances one's ability to solve complex calculus problems efficiently. Learning to evaluate both functions and their derivatives accurately lays the foundation for mastering calculus.