Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ r(x)=3 x^{2}+4 $$

Step-by-Step Solution

Answer

The derivative of \( r(x) = 3x^2 + 4 \) is \( f'(x) = 6x \).

Step 1: Define the Function

Given the function \( r(x) = 3x^2 + 4 \), identify \( f(x) \) as \( 3x^2 + 4 \). Our goal is to find the derivative \( f'(x) \).

Step 2: Set Up the Difference Quotient

Substitute \( f(x) = 3x^2 + 4 \) into the difference quotient formula for the derivative: \[ f'(x) = \lim_{h \to 0} \frac{[(3(x+h)^2 + 4) - (3x^2 + 4)]}{h} \].

Step 3: Expand and Simplify the Numerator

First, expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \). Substitute back:\[ 3((x+h)^2) = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2 \].Plug this into the difference quotient:\[ f'(x) = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 + 4) - (3x^2 + 4)}{h} \].

Step 4: Cancel Terms in the Numerator

Simplify the numerator:\[ (3x^2 + 6xh + 3h^2 + 4) - (3x^2 + 4) = 6xh + 3h^2 \].The expression becomes: \[ f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2}{h} \].

Step 5: Factor and Simplify the Expression

Factor out \( h \) from the numerator:\[ 6xh + 3h^2 = h(6x + 3h) \].So,\[ f'(x) = \lim_{h \to 0} \frac{h(6x + 3h)}{h} \].Cancel the \( h \) terms:\[ f'(x) = \lim_{h \to 0} (6x + 3h) \].

Step 6: Evaluate the Limit

Apply the limit \( h \to 0 \):\[ f'(x) = 6x + 3(0) = 6x \].

Step 7: State the Derivative

Therefore, the derivative of the function \( r(x) = 3x^2 + 4 \) is \( f'(x) = 6x \).

Key Concepts

Difference Quotient

The difference quotient is a foundational step in calculus for finding the derivative of a function. It's all about measuring how a function changes.
For a given function \( f(x) \), the difference quotient is expressed as:

• \( \frac{f(x+h) - f(x)}{h} \)

This formula helps determine the slope of the secant line—which connects two points on the curve—at that interval \( h \).
Here, \( h \) is a small increment from \( x \). As \( h \) approaches zero, the secant line becomes a tangent line, reflecting the function's instantaneous rate of change at point \( x \).
This method is excellent at detailing how small changes in \( x \) affect \( f(x) \). In the given example, \( f(x) = 3x^2 + 4 \), the difference quotient requires evaluating the expression after substituting \( f(x+h) \) and \( f(x) \).

Limit Process

The limit process is crucial in understanding derivatives beyond just formulaic manipulation.
In calculus, the limit helps us solidify our understanding of instantaneous change. For derivatives, we use:

• \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

The "\( \lim_{h \to 0} \)" part is a signal for "getting infinitely close to," which means we're checking how \( f(x+h) \) approaches \( f(x) \) as \( h \) gets smaller and smaller.
In the step-by-step process of taking a derivative, after simplifying the difference quotient, we apply this limit by

• Canceling the \( h \) in the numerator and denominator
• Considering the parts left unaffected as \( h \) shrinks

This results in arriving at the derivative—for our function, \( 6x \) for \( f(x) = 3x^2 + 4 \)—indicating how \( x \) plays a role in the rate of change.

Polynomial Function Differentiation

Differentiating a polynomial function like \( r(x) = 3x^2 + 4 \) is a common task in calculus studies.
Polynomial functions are easy to work with due to their straightforward expressions.
The derivative reflects the rate of change for each of the terms.

• For a term like \( ax^n \), the derivative is \( nax^{n-1} \).
• This is a result of the power rule for differentiation: bring down the exponent as a coefficient, and decrement the exponent by one.

For example, in \( 3x^2 + 4 \):

• The term \( 3x^2 \) becomes \( 6x \) after applying the power rule \((2 \times 3 = 6x^{2-1})\).
• The constant term \( 4 \) vanishes, as constants have a zero rate of change (derivative is zero).

This gives the derivative \( f'(x) = 6x \), capturing how rapidly or slowly the function changes at any point \( x \). The simplifying nature of polynomial derivatives makes them an excellent introduction to differentiation.