Problem 52 - Calculus

Question

Estimate the value of \(f^{\prime}(1)\) by substitutang small values for \(h .\) Then use algebra to find \(f^{\prime}(1)\) exactly.$$f(x)=x^{2}+x+1$$

Step-by-Step Solution

Answer

The exact value of \(f'(1)\) is 3.

Step 1: Understanding the Problem

To estimate the derivative \(f'(1)\), we will use the definition of the derivative: \(f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\). We will first approximate \(f'(1)\) by substituting small values for \(h\) in the difference quotient: \(\frac{{f(1+h) - f(1)}}{h}\).

Step 2: Computing Function Values

Compute \(f(1)\). Since \(f(x) = x^2 + x + 1\), substituting \(x = 1\) gives us:\[f(1) = 1^2 + 1 + 1 = 3.\]

Step 3: Substitution and Approximation

Choose small values of \(h\), such as \(h = 0.1, 0.01, 0.001\) and compute \(f(1+h)\):- For \(h = 0.1:\) \[f(1.1) = (1.1)^2 + 1.1 + 1 = 1.21 + 1.1 + 1 = 3.31\]- For \(h = 0.01:\) \[f(1.01) = (1.01)^2 + 1.01 + 1 = 1.0201 + 1.01 + 1 = 3.0301\]- For \(h = 0.001:\) \[f(1.001) = (1.001)^2 + 1.001 + 1 = 1.002001 + 1.001 + 1 = 3.003001\]

Step 4: Calculating Differences

Now calculate the difference quotient \(\frac{{f(1+h) - f(1)}}{h}\) for each chosen value of \(h\):- For \(h = 0.1:\) \[\frac{{3.31 - 3}}{0.1} = 3.1\]- For \(h = 0.01:\) \[\frac{{3.0301 - 3}}{0.01} = 3.01\]- For \(h = 0.001:\) \[\frac{{3.003001 - 3}}{0.001} = 3.001\]

Step 5: Observing Patterns

Observe that as \(h\) becomes smaller, the approximate derivative values \(3.1, 3.01, 3.001\) get closer to 3. This suggests that the exact value of \(f'(1)\) is 3.

Step 6: Finding the Exact Derivative

Use algebra to find the exact value of \(f'(1)\). The derivative of \(f(x) = x^2 + x + 1\) is found by differentiating each term:\[f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x) + \frac{d}{dx}(1) = 2x + 1 + 0.\]Substitute \(x = 1\) into \(f'(x)\):\[f'(1) = 2(1) + 1 = 3.\]

Step 7: Confirming the Result

The approximation is confirmed by the algebraic calculation. Therefore, the exact value of the derivative \(f'(1)\) is indeed 3.

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus used to estimate the rate of change of a function. It's expressed as the formula \( \frac{{f(x+h) - f(x)}}{h} \) where \( h \) is a very small number. This formula gives an average rate of change of the function \( f(x) \) over the interval \( [x, x+h] \). This method is particularly valuable for approximating derivatives at a point, especially when you don't have immediate access to higher-level calculus concepts.

Using this approach, you obtain multiple estimates for the derivative with smaller and smaller values of \( h \). For instance, by computing the values for \( h = 0.1, 0.01, 0.001 \), one can observe how the average rates of change are approaching a specific value. In this case study, as \( h \) decreases, the difference quotient gives a more precise approximation that converges towards the true derivative at that point.

Limit Definition of Derivative

The limit definition of the derivative is the rigorous foundation of derivative calculation. It is defined as \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \). This formula describes the instantaneous rate of change of the function \( f(x) \) at a specific point \( x \). By taking the limit as \( h \) approaches zero, it allows you to obtain an exact value for the derivative.

This definition emphasizes how a derivative is essentially the slope of the tangent line to the function at a given point. In the exercise, the limit process shows how the approximated difference quotients converge to the actual derivative. By calculating this exact derivative using limits, we achieve a precise slope, removing any approximations involved in earlier steps. It gives a mathematical and accurate method to understand and compute the rate of change.

Polynomial Function

A polynomial function is an expression composed of variables and coefficients, utilizing operations of addition, subtraction, multiplication, and non-negative integer exponents. The function \( f(x) = x^2 + x + 1 \) is a simple polynomial function consisting of three terms: \( x^2 \), \( x \), and a constant \( 1 \).

Polynomials are smooth and continuous functions, which make them pleasant to differentiate using standard rules. Because of their simplicity, the rules of differentiation become straightforward. Their structures allow easy computation of derivatives for each term separately before summing them up. In the exercise, differentiating the polynomial involved applying basic differentiation rules to each part before combining them to find the overall derivative.

Differentiation

Differentiation is the calculus process of finding the derivative of a function. In essence, it is a method to analyze how a function changes at any given point. It utilizes several rules to find derivatives of different types of functions efficiently. These include the power rule, product rule, quotient rule, and chain rule.

• The power rule is frequently used for polynomial differentiation and states that \( \frac{d}{dx}(x^n) = nx^{n-1} \).


• Single-variable polynomials like \( x^2 + x + 1 \) can be differentiated by applying the power rule to each term individually: \( x^2 \) becomes \( 2x \), \( x \) becomes \( 1 \), and constants become zero.


• The process requires identifying the forms of all parts of the function and systematically applying the corresponding differentiation rules.

In practical terms, once you know the rules of differentiation, you can quickly obtain derivatives for wide ranges of functional types, enabling deeper analysis and application of calculus.