Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 7 through 13, find \(f^{\prime}(x), f^{\prime}(0), f^{\prime}(2)\), and \(f^{\prime}(-1) .\) $$ f(x)=3 x+5 $$

Step-by-Step Solution

Answer

The derivative, \(f'(x)\) is \(3\). Evaluating at the points gives \(f'(0) = 3\), \(f'(2) = 3\), and \(f'(-1) = 3\).

Step 1: Calculate Derivative

The derivative of a function represents the rate of change of the function at any point x. Start by taking the derivative \(f'(x)\) of the given function \(f(x) = 3x + 5\). Using basic power rule of differentiation (d/dx[x^n] = n*x^(n-1)), the derivative of \(3x + 5\) becomes \(3\). The derivative of a constant like 5 is zero. So the derivative function \(f'(x)\) will be \(3\). This says that the rate of change of the function \(f(x) = 3x + 5\) is constant and equal to 3.

Step 2: Evaluate at \(x = 0\)

Next, substitute \(x = 0\) into \(f'(x)\). The derivative function is a constant and equals 3, so substituting 0 does not change its value and \(f'(0)\) is also \(3\).

Step 3: Evaluate at \(x = 2\)

Now, substitute \(x = 2\) into \(f'(x)\). Again, since our derivative is a constant, \(f'(2)\) remains equal to \(3\).

Step 4: Evaluate at \(x = -1\)

Lastly, substitute \(x = -1\) into \(f'(x)\). As mentioned before, substituting any value into the derivative function will not change its value because the function is a constant. Thus, \(f'(-1)\) is also \(3\).

Key Concepts

Rate of Change

The concept of the rate of change is essential in understanding how functions behave. When we say "rate of change," we are referring to how much a function's output, or dependent variable, changes based on a change in the input, or independent variable. In the context of derivatives, it tells us how fast and in what direction the function's values are changing at any given point.

• If the rate of change is positive, the function is increasing over that interval.
• If the rate of change is negative, the function is decreasing.
• If it's zero, the function's value is not changing — it's flat.

For linear functions like the given example, the rate of change is constant, represented by the slope of the line. Here, the function is linear:
\( f(x) = 3x + 5 \).
The derivative tells us that the rate of change is constantly 3 everywhere on the line. This means that for every unit increase in \( x \), \( f(x) \) increases by 3.

Power Rule

The Power Rule is a fundamental technique in calculus used to find the derivative of a function. The rule is simple: if you have a term \( x^n \), its derivative is \( n \times x^{n-1} \). This rule is particularly useful for polynomials.
Let's see it in action with the function
\( f(x) = 3x + 5 \).

• The term \( 3x \), applying the Power Rule, has \( n = 1 \). The derivative is simply \( 1 \times 3x^{1-1} = 3 \).
• The term 5 (a constant) can be thought of as \( 5x^0 \), which makes its derivative 0, since \( 0 \times 5x^{-1} = 0 \).

Thus, the derivative
\( f'(x) = 3 \).
This constant derivative indicates a steady rate of increase over the interval.

Differentiation

Differentiation is the process of computing a derivative. It allows us to understand the behavior of functions more deeply by telling us how they change.
Calculating the derivative involves applying rules like the Power Rule to each term in a function. It's a systematic method essential for analyzing and solving real-world problems.
For any given function,
differentiating provides a new function, often noted as \( f'(x) \), that represents the slope at any point on the original curve.

• In our exercise, we differentiated \( f(x) = 3x + 5 \) to find
  \( f'(x) = 3 \).
• This constant derivative confirms that the function is linear with a fixed slope.
• When you substitute any value into this derivative, like \( x = 0, 2, \text{or} -1 \), the slope (and hence the rate of change) remains the same.

Differentiating is like a tool in a mathematician's toolkit that helps reveal a function's underlying tendencies and rates of change.