Problem 47 - Calculus: Early Transcendentals

Question

Suppose \( f \) and \( g \) are continuous functions such that \( g(2) = 6 \) and \( \displaystyle \lim_{x \to 2} [3f(x) + f(x)g(x)] = 36 \). Find \( f(2) \).

Step-by-Step Solution

Answer

\( f(2) = 4 \).

Step 1: Understand the Given Limit Condition

We know from the problem statement that \( \lim_{x \to 2} [3f(x) + f(x)g(x)] = 36 \). Because both \( f \) and \( g \) are continuous functions at \( x = 2 \), we can directly substitute \( x = 2 \) into the expression. This means we can replace \( f(x) \) with \( f(2) \) and \( g(x) \) with \( g(2) \) in the given limit expression.

Step 2: Substitute Known Values

We substitute \( g(2) = 6 \) into the expression. The limit becomes: \[ 3f(2) + f(2) \times 6 = 36 \]. Simplify this expression by combining like terms.

Step 3: Simplify the Expression

Rewrite the equation as \( 3f(2) + 6f(2) = 36 \). By combining like terms, we have \( 9f(2) = 36 \). This simplifies to show the combined influence of \( f(2) \) across the expression.

Step 4: Solve for \( f(2) \)

To isolate \( f(2) \), divide both sides of the equation \( 9f(2) = 36 \) by 9. This gives \( f(2) = \frac{36}{9} = 4 \). So, \( f(2) = 4 \).

Key Concepts

Continuous functions

A continuous function is an essential concept in calculus and graph analysis. To say a function is continuous at a point means it behaves as expected, with no jumps or interruptions. For a function \( f(x) \) to be continuous at a point \( x = c \), the following three conditions must hold:

• The function \( f(x) \) is defined at \( x = c \), meaning \( f(c) \) exists.
• The limit of the function as \( x \) approaches \( c \) exists, denoted as \( \lim_{x \to c} f(x) \).
• The value of the function at that point equals the limit, so \( f(c) = \lim_{x \to c} f(x) \).

Continuous functions allow for straightforward computation of limits because the values of the functions can be directly substituted at points of interest. This makes analyzing behavior at specific points much simpler and less prone to error.

Limits in calculus

Limits are foundational to calculus and help us understand the behavior of functions as they approach specific points. The notation \( \lim_{x \to a} f(x) = L \) implies that as \( x \) gets closer and closer to \( a \), the function \( f(x) \) gets closer to \( L \).

Limits are useful to:

• Predict the value a function approaches as the input approaches some value.
• Define and analyze the continuity of a function.
• Determine the value of a function at points that are problematic, like dividing by zero or points not in the function's domain.

In the given exercise, we used limits to analyze the function \( 3f(x) + f(x)g(x) \) as \( x \) approaches 2. This enables us to substitute directly into the expression using continuity, helping solve for \( f(2) \).

Function substitution

Function substitution is a handy tool in calculus used to replace variables with known or simpler values to make problem-solving streamlined. It allows us to solve complex expressions by substituting points of interest directly into equations, assuming continuity exists.

• When faced with functions like \( f(x) \) or \( g(x) \), substitution lets us directly use known values like \( f(c) \) or \( g(c) \) when the functions are continuous.
• This process often simplifies solving for unknowns, such as the evaluation of limits and derivatives.

In the original exercise, by substituting \( g(2) = 6 \) and using the continuity of \( f(x) \) and \( g(x) \), the expression simplified to a straightforward algebraic equation: \( 9f(2) = 36 \). This substitution led to the solution of finding \( f(2) \).