Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

For the function, \(F(x)=x^{2}+x,\) compute the following (a) \(\frac{F(5)-F(3)}{5-3}\) (b) \(\frac{F(3+2)-F(3)}{2}\) (c) \(\frac{F(b)-F(a)}{b-a}\) (d) \(\frac{F(a+h)-F(a)}{h}\)

Step-by-Step Solution

Answer

(a) 9 (b) 9 (c) b+a+1 (d) 2a+h+1

Step 1: Evaluate F(5) and F(3) for Part (a)

First, compute \( F(5) \) and \( F(3) \). Given \( F(x) = x^2 + x \), substitute 5 for \( x \): \( F(5) = 5^2 + 5 = 25 + 5 = 30 \). Similarly, compute \( F(3) = 3^2 + 3 = 9 + 3 = 12 \).

Step 2: Compute Expression for Part (a)

Now find \( \frac{F(5) - F(3)}{5 - 3} \). Substitute the values found: \( \frac{30 - 12}{2} = \frac{18}{2} = 9 \).

Step 3: Evaluate F(3+2) and F(3) for Part (b)

Next, for part (b), compute \( F(3 + 2) \) and \( F(3) \) which was already calculated. So, \( F(5) = 30 \) from earlier. \( F(3) = 12 \).

Step 4: Compute Expression for Part (b)

Calculate \( \frac{F(3+2) - F(3)}{2} \) using earlier results: \( \frac{30 - 12}{2} = 9 \).

Step 5: Use the Function to Formulate General Expression for Part (c)

Use the formula \( F(x) = x^2 + x \) with \( b \) and \( a \) to find \( F(b) = b^2 + b \) and \( F(a) = a^2 + a \).

Step 6: Compute Expression for Part (c)

Substitute into \( \frac{F(b) - F(a)}{b - a} \): \( \frac{b^2 + b - (a^2 + a)}{b - a} = \frac{(b^2 - a^2) + (b-a)}{b - a} \). Use factoring: \( \frac{(b-a)(b+a) + (b-a)}{b-a} \), which simplifies to \( b+a+1 \) since \( b-a \) cancels out.

Step 7: Use the Function to Formulate General Expression for Part (d)

For part (d), find \( F(a+h) = (a+h)^2 + (a+h) \). This expands to \( a^2 + 2ah + h^2 + a + h \).

Step 8: Compute Expression for Part (d)

Substitute into \( \frac{F(a+h) - F(a)}{h} \) using \( F(a) = a^2 + a \): \( \frac{a^2 + 2ah + h^2 + a + h - a^2 - a}{h} = \frac{2ah + h^2 + h}{h} \). Simplify this to \( 2a + h + 1 \) after cancelling \( h \).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in differential calculus. It forms the basis for understanding many derivative calculations. The general formula for the difference quotient is given by:\[ \frac{F(x+h) - F(x)}{h} \]For our specific exercise, the difference quotient is applied in parts (a), (b), (c), and (d) to evaluate changes in the function's value over different intervals.

• Part (a): The computation involves an interval between two specific points, 3 and 5. This required finding \( F(5) \) and \( F(3) \) and then calculating the difference quotient to be 9.
• Part (b): With additions in the input (3 and 3+2), the difference quotient highlighted the change over a segment, resulting again in a value of 9.
• Part (c): Here, the difference quotient was used in a general sense with inputs \( a \) and \( b \), leading to a formula \( b + a + 1 \).
• Part (d): This step generalizes the idea further to inputs \( a \) and \( a + h \), reflecting a more abstract application with a result of \( 2a + h + 1 \).

Understanding this quotient is crucial, as it paves the way to defining the derivative as \( h \) approaches zero.

Slope of Secant Line

The slope of a secant line offers insights into the average rate of change of a function over an interval and is computed using the difference quotient. For the function \( F(x) \), we're looking into how its values change between points. In simple terms, the secant line connects two points on the curve of the function. This line's slope represents the average rate at which the function \( F(x) \) changes between these two inputs.

• When you calculate \( \frac{F(5) - F(3)}{2} \) in part (a), you're finding the slope of the secant line connecting \( F(3) \) and \( F(5) \).
• Similarly, in part (b), \( \frac{F(3+2) - F(3)}{2} \) reflects the slope of a secant line for specific inputs closer to each other. Despite different intervals, both parts yield the same slope of 9, indicating consistent change rates.

By comparing these slopes across various parts, students can begin to appreciate how function changes are represented geometrically.

Function Evaluation

Function evaluation is the process of finding the output of a function for specific input values. In our exercise, we evaluated the function \( F(x) = x^2 + x \) at different points, such as 3, 5, and abstract points like \( a \) and \( b \).Here’s how we conducted function evaluations:

• Substituting values: For part (a), compute \( F(5) = 5^2 + 5 \) and \( F(3) = 3^2 + 3 \) to receive 30 and 12, respectively.
• General expression: For points like \( a \) and \( b \), function evaluations led to expressions like \( F(a) = a^2 + a \) and \( F(b) = b^2 + b \).

These calculations form the basis for more complex operations, such as finding the difference quotient and understanding how changes in input values affect the function's output.

Algebraic Simplification

Algebraic simplification is essential for making expressions more manageable and easier to interpret. This step often involves factoring, cancelling terms, and combining like terms.In solving parts (c) and (d), simplifications were key:

• Part (c): We started with \( \frac{b^2 + b - (a^2 + a)}{b-a} \). By factoring it as \( \frac{(b-a)(b+a) + (b-a)}{b-a} \), we could simplify to \( b + a + 1 \), effectively canceling \( b-a \).
• Part (d): Simplification involved expanding and simplifying \( \frac{a^2 + 2ah + h^2 + a + h - a^2 - a}{h} \) to \( 2a + h + 1 \), by cancelling out \( h \). This demonstrated algebra's role in streamlining function-based expressions.

These simplifications not only make the expressions easier to work with but also reveal deeper insights into the behavior and characteristics of functions.