Problem 34 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. $$ f(x)=3(5-x)^{2} \quad(5,0) $$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point (5,0) is 0.

Step 1: Apply the chain rule to find the derivative

The derivative of \( f(x) = 3(5-x)^2 \) using the chain rule can be found as follows: First, find the derivative of the outside function and leave the inside function as it is. The outside function here is \( 3(u)^2 \), so its derivative is \( 2*3*u = 6u \). Replace \( u \) with the inside function \( (5-x) \) to get \( 6(5-x) \). Then, multiply by the derivative of the inside function. As the derivative of \( (5-x) \) is \( -1 \), the derivative of \( f(x) \) is \( 6(5-x)*(-1) = -6(5-x) \).

Step 2: Simplify the derivative expression

After expanding and simplifying, the derivative \( f'(x) = -6(5-x) \) yields \( f'(x) = -30 + 6x \).

Step 3: Evaluate the derivative at the given point

Finally, substitute the x coordinate of the given point into the derivative. For point (5,0), replace \( x \) with 5 in \( f'(x) = -30 + 6x \), to get \( f'(5) = -30 + 6*5 = 0 \).

Key Concepts

Derivative

A derivative helps you understand how a function changes at any given point. When dealing with derivative, you often look for the rate at which a function is increasing or decreasing. The derivative represents the slope of a tangent line at a particular point on the function's curve.

In our exercise, the function is given as:

• \( f(x) = 3(5-x)^2 \)

To find the derivative of this function, we use the **chain rule**, a tool used when taking derivatives of composite functions. Here, the outer function is \( 3(u)^2 \) and the inner function is \( (5-x) \). The chain rule states that you take the derivative of the outer function first while keeping the inner function intact, and then multiply it by the derivative of the inner function.

For the outer function \( 3(u)^2 \), its derivative becomes \( 6u \). Then, substitute back \( u = (5-x) \) giving us \( 6(5-x) \). Next, we multiply by the derivative of the inner function which is \(-1\). Thus, the full derivative at this point is:

• \( f'(x) = -6(5-x) \)

Understanding how parts of a function influence its derivative helps you analyze and predict behavior of functions efficiently.

Function Evaluation

Evaluating a function means substituting a value for a variable into the function. This process helps determine the specific output or result based on given inputs.

For our function \( f(x) \), after finding its derivative \( f'(x) = -6(5-x) \), the next step is to evaluate this expression at a specific point. We use the point \((5,0)\) provided in the problem. This means we're substituting \( x = 5 \) into the expression we've derived for the function's rate of change:

By replacing \( x \) with 5, the expression becomes:

• \( f'(5) = -6(5-5) \)
• Simplifying, \( f'(5) = 0 \)

The evaluated derivative \( f'(5) = 0 \) shows that at \( x=5 \), the function has a slope of zero. This indicates a special point — possibly a peak or a flat point on the graph of the function. Evaluating derivatives like this is key when exploring how functions behave at specific instances.

Simplifying Expressions

Simplifying expressions is a crucial step in solving math problems efficiently. It involves reducing expressions to their simplest forms, making them easier to work with and understand.

In this problem, after using the chain rule, we derived the following expression for the derivative of the function:

• \( f'(x) = -6(5-x) \)

To simplify:

• Expand: \( -6*5 + 6*x \)
• This results in \( -30 + 6x \)

Breaking down expressions into simpler forms can help reveal important characteristics of the function, such as intercepts and slopes. Simplification makes it easier to substitute specific values, such as during function evaluation. It's an essential technique that provides clarity and can sometimes even simplify subsequent calculations in your overall solution strategy.