Problem 7 - Active Calculus

Question

Consider the function \(g(x)=x^{2}-x+3\). a. Use the limit definition of the derivative to determine a formula for \(g^{\prime}(x)\). b. Use a graphing utility to plot both \(y=g(x)\) and your result for \(y=g^{\prime}(x) ;\) does your formula for \(g^{\prime}(x)\) generate the graph you expected? c. Use the limit definition of the derivative to find a formula for \(p^{\prime}(x)\) where \(p(x)=5 x^{2}-\) \(4 x+12\). d. Compare and contrast the formulas for \(g^{\prime}(x)\) and \(p^{\prime}(x)\) you have found. How do the constants \(5,4,12,\) and 3 affect the results?

Step-by-Step Solution

Answer

The derivatives are \( g'(x) = 2x - 1 \) and \( p'(x) = 10x - 4 \). The graph of \( g'(x) \) matches the expected slope behavior. Higher constants in \( p(x) \) result in a steeper slope for \( p'(x) \).

Step 1: Define the Limit Definition of the Derivative

The limit definition of the derivative is given by \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]

Step 2: Apply the Definition for g(x)

For the function \( g(x) = x^2 - x + 3 \), \[ g'(x) = \lim_{{h \to 0}} \frac{{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}}{h} \]

Step 3: Expand and Simplify the Expression

Expand \((x+h)^2\) to get \( x^2 + 2xh + h^2\). Plug this back into the limit: \[ g'(x) = \lim_{{h \to 0}} \frac{{x^2 + 2xh + h^2 - x - h + 3 - x^2 + x - 3}}{h} \] Combine like terms: \[ g'(x) = \lim_{{h \to 0}} \frac{{2xh + h^2 - h}}{h} \]

Step 4: Factor Out h and Simplify

Factor out \( h \) from the numerator: \[ g'(x) = \lim_{{h \to 0}} \frac{{h(2x + h - 1)}}{h} \] Cancel the \( h \): \[ g'(x) = \lim_{{h \to 0}} (2x + h - 1) \] Take the limit as \( h \) approaches 0: \[ g'(x) = 2x - 1 \]

Step 5: Graphing g(x) and g'(x)

Using a graphing utility, plot \( y = x^2 - x + 3 \) and \( y = 2x - 1 \). Verify that the derivative graph matches the slope behavior of \( g(x) \).

Step 6: Apply the Limit Definition for p(x)

For the function \( p(x) = 5x^2 - 4x + 12 \), \[ p'(x) = \lim_{{h \to 0}} \frac{{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}}{h} \]

Step 7: Expand and Simplify the Expression for p(x)

Expand \((x+h)^2\) and simplify: \[ p'(x) = \lim_{{h \to 0}} \frac{{5(x^2 + 2xh + h^2) - 4x - 4h + 12 - 5x^2 + 4x - 12}}{h} \] Combine like terms: \[ p'(x) = \lim_{{h \to 0}} \frac{{10xh + 5h^2 - 4h}}{h} \]

Step 8: Factor Out h and Simplify

Factor out \( h \): \[ p'(x) = \lim_{{h \to 0}} \frac{{h(10x + 5h - 4)}}{h} \] Cancel the \( h \): \[ p'(x) = \lim_{{h \to 0}} (10x + 5h - 4) \] Take the limit as \( h \) approaches 0: \[ p'(x) = 10x - 4 \]

Step 9: Compare g'(x) and p'(x)

The formulas are \( g'(x) = 2x - 1 \) and \( p'(x) = 10x - 4 \). The constants 5, 4, 12, and 3 affect the derivatives in terms of their coefficients and constants.

Key Concepts

derivative formula

In calculus, the derivative formula is crucial as it defines the rate at which a function is changing at any given point. The most fundamental definition is given by the limit process:
\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
This formula essentially measures the slope of the tangent line to the curve of a function at a particular point. By letting \( h \) approach zero, we get the exact instantaneous rate of change. This concept is foundational for more advanced topics in calculus, such as optimization and motion analysis.

function differentiation

Function differentiation is the process of finding the derivative of a function. For example, let's differentiate the function \( g(x) = x^2 - x + 3 \) using the limit definition of derivatives:
\[ g'(x) = \lim_{{h \to 0}} \frac{{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}}{h} \]
This involves expanding \((x+h)^2\) to \(x^2 + 2xh + h^2\) and simplifying the expression:
\[ g'(x) = \lim_{{h \to 0}} \frac{{2xh + h^2 - h}}{h} \]
Factor out and cancel \( h \):
\[ g'(x) = \lim_{{h \to 0}} (2x + h - 1) = 2x - 1 \]
Through differentiation, we find that the derivative, \( g'(x) = 2x - 1 \), represents the slope of the original function at any point \( x \).

calculus problem-solving

Solving problems in calculus, especially those involving derivatives, requires systematic steps. For instance, let's tackle a problem using the limit definition for another function \( p(x) = 5x^2 - 4x + 12 \):
\[ p'(x) = \lim_{{h \to 0}} \frac{{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}}{h} \]
Expand \((x+h)^2\) and combine terms:
\[ p'(x) = \lim_{{h \to 0}} \frac{{10xh + 5h^2 - 4h}}{h} \]
Factor out \( h \) and simplify:
\[ p'(x) = \lim_{{h \to 0}} (10x + 5h - 4) = 10x - 4 \]
By following these steps, we ensure accuracy and clarity in problem-solving. Compare this process to the differentiation of \( g(x) \) to appreciate how constants affect the derivative.

graphing derivatives

Graphing derivatives helps visualize how a function changes. For example, consider \( g(x) = x^2 - x + 3 \) and its derivative \( g'(x) = 2x - 1 \). By using a graphing utility:
- Plot \( y = g(x) \) and \( y = g'(x) \)
- Observe how the slope of \( g(x) \) matches the values of \( g'(x) \)
For instance, when \( x \) increases, the slope of \( g(x) \) becomes steeper, reflected in the increasing value of \( 2x - 1 \).
Similarly, for \( p(x) = 5x^2 - 4x + 12 \) and \( p'(x) = 10x - 4 \), graphing these can reveal how the steeper quadratic function has a rapidly changing slope mirrored in its derivative. This visual approach enhances understanding of the relationship between a function and its derivative.