Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

\(\lim _{x \rightarrow a} \frac{F(x)-F(a)}{x-a}\) a. \(\quad F(x)=x^{2} \quad a=-2\) b. \(\quad F(x)=17 \quad a=0\) c. \(\quad F(x)=2 x^{3} \quad a=2\) d. \(\quad F(x)=x^{2}+2 x \quad a=1\) e. \(\quad F(x)=\frac{1}{x} \quad a=\frac{1}{2}\) f. \(\quad F(x)=3 x^{2}-5 x \quad a=7\) g. \(\quad F(x)=3 \sqrt{x} \quad a=4\) h. \(\quad F(x)=x^{2}+2 x+1 \quad a=-1\) i. \(F(x)=\frac{4}{x}+5 \quad a=2 \quad\) j. \(\quad F(x)=x^{6} \quad a=2\) k. \(\quad F(x)=\frac{1}{x^{3}} \quad a=2\) l. \(F(x)=x^{10}\) \(a=2\) m. \(\quad F(x)=\frac{4}{x^{5}} \quad a=2\) n. \(\quad F(x)=x^{67} \quad a=1\)

Step-by-Step Solution

Answer

a. -4, b. 0, c. 24, d. 4, e. -4, f. 37, g. 3/4, h. 0, i. -1, j. 192, k. -3/16, l. 5120, m. -5/16, n. 67.

Step 1: Calculate the Derivative

The given exercise can be tackled using the definition of the derivative. The derivative of a function at a point \(a\) is defined as \(F'(a) = \lim_{x \rightarrow a} \frac{F(x) - F(a)}{x-a}\). Calculating this limit gives us the derivative of the function \(F(x)\) evaluated at \(x = a\). We'll apply this definition to each sub-problem.

Step 2a: Solve \(F(x) = x^2\), \(a = -2\)

Find the derivative: \(F'(x) = 2x\). Evaluate it at \(x = -2\): \[F'(-2) = 2(-2) = -4\].

Step 2b: Solve \(F(x) = 17\), \(a = 0\)

The derivative of a constant is zero: \(F'(x) = 0\). Thus, \(F'(0) = 0\).

Step 2c: Solve \(F(x) = 2x^3\), \(a = 2\)

Find the derivative: \(F'(x) = 6x^2\). Evaluate it at \(x = 2\): \[F'(2) = 6(2)^2 = 24\].

Step 2d: Solve \(F(x) = x^2 + 2x\), \(a = 1\)

Find the derivative: \(F'(x) = 2x + 2\). Evaluate it at \(x = 1\): \[F'(1) = 2(1) + 2 = 4\].

Step 2e: Solve \(F(x) = \frac{1}{x}\), \(a = \frac{1}{2}\)

Find the derivative using the power rule: \(F'(x) = -\frac{1}{x^2}\). Evaluate it at \(x = \frac{1}{2}\): \[F'(\frac{1}{2}) = -\frac{1}{(\frac{1}{2})^2} = -4\].

Step 2f: Solve \(F(x) = 3x^2 - 5x\), \(a = 7\)

Find the derivative: \(F'(x) = 6x - 5\). Evaluate it at \(x = 7\): \[F'(7) = 6(7) - 5 = 37\].

Step 2g: Solve \(F(x) = 3\sqrt{x}\), \(a = 4\)

First rewrite \(F(x) = 3x^{1/2}\). Find the derivative: \(F'(x) = \frac{3}{2}x^{-1/2}\). Evaluate it at \(x = 4\): \[F'(4) = \frac{3}{2\sqrt{4}} = \frac{3}{4}\].

Step 2h: Solve \(F(x) = x^2 + 2x + 1\), \(a = -1\)

Find the derivative: \(F'(x) = 2x + 2\). Evaluate it at \(x = -1\): \[F'(-1) = 2(-1) + 2 = 0\].

Step 2i: Solve \(F(x) = \frac{4}{x} + 5\), \(a = 2\)

Find the derivative: \(F'(x) = -\frac{4}{x^2}\). Evaluate it at \(x = 2\): \[F'(2) = -\frac{4}{2^2} = -1\].

Step 2j: Solve \(F(x) = x^6\), \(a = 2\)

Find the derivative: \(F'(x) = 6x^5\). Evaluate it at \(x = 2\): \[F'(2) = 6(2)^5 = 192\].

Step 2k: Solve \(F(x) = \frac{1}{x^3}\), \(a = 2\)

Find the derivative using the power rule: \(F'(x) = -3x^{-4} = -\frac{3}{x^4}\). Evaluate it at \(x = 2\): \[F'(2) = -\frac{3}{2^4} = -\frac{3}{16}\].

Step 2l: Solve \(F(x) = x^{10}\), \(a = 2\)

Find the derivative: \(F'(x) = 10x^9\). Evaluate it at \(x = 2\): \[F'(2) = 10(2)^9 = 5120\].

Step 2m: Solve \(F(x) = \frac{4}{x^5}\), \(a = 2\)

Find the derivative: \(F'(x) = -20x^{-6} = -\frac{20}{x^6}\). Evaluate it at \(x = 2\): \[F'(2) = -\frac{20}{2^6} = -\frac{20}{64} = -\frac{5}{16}\].

Step 2n: Solve \(F(x) = x^{67}\), \(a = 1\)

Find the derivative: \(F'(x) = 67x^{66}\). Evaluate it at \(x = 1\): \[F'(1) = 67(1)^{66} = 67\].

Key Concepts

Calculus

Calculus is a fascinating area of mathematics that helps us understand how functions change. It's like a toolset for understanding motion and change. In calculus, there are two main areas: differentiation and integration. Differentiation helps us find rates of change or the slope of curves, while integration gives us the total accumulation of quantities.

When you're looking at a curve and want to know how steep it is at any given point, differentiation is the process you'll use. By working with derivatives, which are a core part of calculus, you can determine exactly how a function behaves. Comprehending calculus can be incredibly useful in many real-world applications, such as physics, engineering, and economics, where recognizing change or optimizing conditions is crucial.

Broadly, calculus allows us to work with abstract concepts in a practical way, giving us the ability to model real-world phenomena with accuracy and precision. Understanding calculus opens up a whole new language for exploring patterns and solving complex problems.

Limit Definition of Derivative

The limit definition of a derivative is foundational in calculus. It gives us a formal way to find the derivative of a function at a particular point. The derivative tells us the rate of change or the slope of the function at that point. The definition is expressed as:\[F'(a) = \lim_{x \rightarrow a} \frac{F(x) - F(a)}{x-a}\]This might seem complex, but it's essentially calculating how much the function's value changes as we make very small changes to its input values (as \(x\) approaches \(a\)). This process of taking limits helps us find the exact slope of a tangent line to the curve at any point.

To apply it, substitute your function into the formula, rearrange to simplify, and calculate the limit as \(x\) approaches \(a\). This process is critical for finding derivatives of any function, especially when these functions cannot be easily differentiated using standard rules.

Function Evaluation

Function evaluation is the process of determining the output of a function for specific input values. In the context of derivatives, once you've found the general derivative of a function, evaluating it for a specific value gives you the instantaneous rate of change at that point.

For example, after determining the derivative of a function like \(F(x)=x^2\), you would evaluate it at a certain \(x\) value, say \(x=-2\), to get \(F'(-2) = 2(-2) = -4\). This means that the slope of the function at \(x = -2\) is -4, giving a detailed insight into how the function behaves locally.

Function evaluation helps translate abstract mathematical calculations into tangible information, which is crucial in scenarios like estimating future values or adjusting parameters for optimal outcomes.

Mathematical Modeling

Mathematical modeling involves using mathematical language and structures to represent real-world problems. It's like building a bridge between abstract math and practical applications. In calculus, derivatives play a significant role in modeling because they provide precise rates of change and are crucial for predicting behavior, optimizing solutions, and understanding dynamic systems.

By constructing a model using calculus, you can simulate and analyze scenarios ranging from natural phenomena like population growth to engineering systems like the control of a vehicle. The ability to model mathematically allows us to anticipate outcomes, evaluate solutions, and enhance the efficiency of systems.
For instance, in a business environment, modeling might help you decide how changing prices could affect demand. In engineering, it could be used to determine how varying forces affect stress on a structure. Understanding mathematical modeling can open up numerous opportunities to apply mathematical insights to solve practical problems.