Problem 1 - University Calculus: Early Transcendentals

Question

Using the definition, calculate the derivatives of the functions in Exercises \(1-6 .\) Then find the values of the derivatives as specified. $$f(x)=4-x^{2} ; \quad f^{\prime}(-3), f^{\prime}(0), f^{\prime}(1)$$

Step-by-Step Solution

Answer

The derivatives are: \( f'(-3) = 6 \), \( f'(0) = 0 \), \( f'(1) = -2 \).

Step 1: Understand the Problem

The task is to find the derivative of the function \( f(x) = 4 - x^2 \) using the definition of the derivative, and then evaluate the derivative at the specified points: \( x = -3 \), \( x = 0 \), and \( x = 1 \).

Step 2: Use the Definition of the Derivative

The derivative of a function \( f(x) \) at a point \( x = a \) is defined as \( f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h} \). We apply this to find a general expression for \( f'(x) \).

Step 3: Apply the Function to the Definition

Substitute \( f(x) = 4 - x^2 \) into the definition of the derivative:\[ f'(x) = \lim_{{h \to 0}} \frac{(4 - (x+h)^2) - (4 - x^2)}{h} \]

Step 4: Simplify the Expression in the Numerator

Expand \((x+h)^2\) to get \(x^2 + 2xh + h^2\). Substitute back into the expression:\[ f'(x) = \lim_{{h \to 0}} \frac{4 - (x^2 + 2xh + h^2) - 4 + x^2}{h} \]

Step 5: Simplify Further

Combine like terms in the numerator to get:\[ f'(x) = \lim_{{h \to 0}} \frac{-2xh - h^2}{h} \]

Step 6: Factor and Simplify

Factor out \(h\) in the numerator:\[ f'(x) = \lim_{{h \to 0}} \frac{h(-2x - h)}{h} = \lim_{{h \to 0}} (-2x - h) \]Now, as \(h\) approaches 0, the expression simplifies to \(-2x\).

Step 7: Conclusion of the Derivative

The derivative of the function \( f(x) = 4 - x^2 \) is \( f'(x) = -2x \).

Step 8: Evaluate the Derivative at \( x = -3 \)

Substitute \( x = -3 \) into \( f'(x) = -2x \): \[ f'(-3) = -2(-3) = 6 \]

Step 9: Evaluate the Derivative at \( x = 0 \)

Substitute \( x = 0 \) into \( f'(x) = -2x \): \[ f'(0) = -2(0) = 0 \]

Step 10: Evaluate the Derivative at \( x = 1 \)

Substitute \( x = 1 \) into \( f'(x) = -2x \): \[ f'(1) = -2(1) = -2 \]

Key Concepts

Definition of derivative

The concept of a derivative is fundamental in calculus. Essentially, it measures the rate at which a function changes at any given point. If you think about a car driving down a road, the derivative would be similar to the speedometer reading, showing your speed at a specific moment.
To find this instantaneous rate of change for a function \( f(x) \), we use the definition of the derivative, which is formally written as:

• \( f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h} \)

This formula looks at a small change around the point \( a \), and calculates how much the function \( f(x) \) changes relative to that small change. The \( \lim_{{h \to 0}} \) part means that we want to find the limit as \( h \), an arbitrarily tiny number, approaches 0.
In practical terms for our exercise, this definition helps us calculate the derivative of the function \( f(x) = 4 - x^2 \) step by step, showing how changes in \( x \) affect the output of the function.

Function differentiation

Differentiation is the process of applying rules and formulas to determine the derivative of a function. For the function \( f(x) = 4 - x^2 \), we want to find its derivative \( f'(x) \).
We use the steps outlined in the definition of the derivative:

• Start by substituting \( f(x) \) into the definition: \( \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \)
• For \( f(x) = 4 - x^2 \), this becomes \( \lim_{{h \to 0}} \frac{(4 - (x+h)^2) - (4 - x^2)}{h} \)

By expanding and simplifying the expression \( (x+h)^2 \) as \( x^2 + 2xh + h^2 \), and substituting back, we further simplify:

• \( \lim_{{h \to 0}} \frac{-2xh - h^2}{h} \)
• After factoring out \( h \), it becomes \( \lim_{{h \to 0}} (-2x - h) \)

As \( h \) approaches 0, the expression simplifies to \( -2x \). This process shows us how the differentiation rules neatly lead us to conclude that the derivative, or the function's rate of change, is \( f'(x) = -2x \).

Evaluating derivatives at specific points

Now that we have the general derivative of \( f(x) = 4 - x^2 \), which is \( f'(x) = -2x \), we can evaluate this at specific points. This means substituting different values of \( x \) into our derived function to find the rate of change at those points.
Let's take three examples:

• For \( x = -3 \): Substitute into the derivative function to get \( f'(-3) = -2(-3) = 6 \). This tells us that at \( x = -3 \), the function is increasing at a rate of 6.
• For \( x = 0 \): Substitute to receive \( f'(0) = -2(0) = 0 \). The derivative at this point is 0, indicating no change or a flat slope.
• For \( x = 1 \): Substituting gives us \( f'(1) = -2(1) = -2 \). Here, the function is decreasing at a rate of 2.

These evaluations provide insight into how the function behaves at these specific points, whether it's increasing, decreasing, or remaining constant.