Problem 58 - Brief Applied Calculus

Question

An automobile dealership finds that the number of cars that it sells on day \(x\) of an advertising campaign is \(S(x)=-x^{2}+10 x\) (for \(0 \leq x \leq 7)\) a. Find \(S^{\prime}(x)\) by using the definition of the derivative. b. Use your answer to part (a) to find the instantaneous rate of change on day \(x=3\). c. Use your answer to part (a) to find the instantaneous rate of change on day \(x=6\).

Step-by-Step Solution

Answer

a. \( S'(x) = -2x + 10 \); b. 4 cars/day; c. -2 cars/day.

Step 1: Define the Function and its Derivative

The given function for the number of cars sold is \( S(x) = -x^2 + 10x \). We want to find \( S'(x) \) using the definition of the derivative. The definition states that the derivative \( S'(x) \) is given by the limit: \[ S'(x) = \lim_{h \to 0} \frac{S(x+h) - S(x)}{h} \]

Step 2: Expand \(S(x+h)\)

Using the definition, expand \( S(x+h) \):\[ S(x+h) = -(x+h)^2 + 10(x+h) \]This expands to:\[ S(x+h) = -(x^2 + 2xh + h^2) + 10x + 10h \]So:\[ S(x+h) = -x^2 - 2xh - h^2 + 10x + 10h \]

Step 3: Compute the Difference \(S(x+h) - S(x)\)

Now, compute the difference:\[ S(x+h) - S(x) = (-x^2 - 2xh - h^2 + 10x + 10h) - (-x^2 + 10x) \]Simplify this to:\[ = -2xh - h^2 + 10h \]

Step 4: Find the Limit as \( h \to 0 \)

Use the limit definition to find the derivative:\[ S'(x) = \lim_{h \to 0} \frac{-2xh - h^2 + 10h}{h} \]Factor \( h \) out of the numerator:\[ S'(x) = \lim_{h \to 0} \frac{h(-2x - h + 10)}{h} \]Cancel \( h \) and compute the limit:\[ S'(x) = \lim_{h \to 0} (-2x - h + 10) = -2x + 10 \]

Step 5: Calculate Instantaneous Rate of Change at \( x = 3 \)

Using \( S'(x) = -2x + 10 \), substitute \( x = 3 \) to find:\[ S'(3) = -2(3) + 10 = -6 + 10 = 4 \]This means the instantaneous rate of change at \( x = 3 \) is 4 cars per day.

Step 6: Calculate Instantaneous Rate of Change at \( x = 6 \)

Again using \( S'(x) = -2x + 10 \), substitute \( x = 6 \) to find:\[ S'(6) = -2(6) + 10 = -12 + 10 = -2 \]This means the instantaneous rate of change at \( x = 6 \) is -2 cars per day.

Key Concepts

Instantaneous Rate of Change

When you think about the instantaneous rate of change, it's like catching a car's speedometer as it shows how fast you're going at a specific moment. It tells you how quickly something is changing at a specific point in time, rather than over an entire interval. In the context of our car dealership, this means finding out how many cars are being sold per day at any given day within their advertising campaign.
To find this rate of change for the sales function during the campaign, we use the derivative of the function of sales. If on day 3 we find that the dealership sells cars at a rate of 4 cars per day, it means this pace only applies at precisely that moment. It doesn't predict past or future sales. Instead, it offers a snapshot based on current factors.
This precise calculation requires derivatives, acting as a tool to zoom in on the function's behavior at an exact day of the campaign.

Definition of Derivative

The derivative is one of the most fundamental tools in calculus. It's like a mathematician's magnifying glass, allowing us to see how a function behaves at the tiniest levels of change. Derivatives help us understand how things change instantaneously.
The classic definition uses a limit approach: \[ S'(x) = \lim_{h \to 0} \frac{S(x+h) - S(x)}{h} \] Here, \(h\) represents a very small change in \(x\), and the limit describes what happens as \(h\) gets closer and closer to zero. This formula is used to figure out how the function \(S(x) = -x^2 + 10x\) changes. By plugging in values, we determine the rate at any given day during the advertising period.
So in our example, derivative \(S'(x)\) describes how the dealership's car sales change day by day, giving us insights into advertising effectiveness.

Polynomial Function Derivatives

When dealing with polynomial functions, derivatives become more predictable and manageable. A polynomial like our function \(S(x) = -x^2 + 10x\) consists of terms like \(-x^2\) and \(10x\). Each of these parts is simple to deal with using basic rules of differentiation.
To find the derivative of \(S(x)\), you apply power rules:

• Power rule: For any term \(ax^n\), the derivative is \(n \cdot ax^{n-1}\).

So for \(-x^2\), the derivative results in \(-2x^1\) (or just \(-2x\)), and for \(10x\), it simplifies to \(10\), since you're essentially reverting to the coefficient alone.
Consequently, the derivative of our function becomes \(-2x + 10\), a linear expression that makes it easy to assess changes in car sales as a function of \(x\), the day of the campaign. This simplification shows how derivatives streamline working with polynomial functions, giving us quick insights into dynamic changes without complex calculations.