Problem 3 - Calculus and its applications

Question

For \(f(x, y)=3^{x}+7 x y,\) find \(f(0,-2), f(-2,1),\) and \(f(2,1)\).

Step-by-Step Solution

Answer

\( f(0, -2) = 1; f(-2, 1) = \frac{-125}{9}; f(2, 1) = 23 \).

Step 1: Understand the Function

The given function is \( f(x, y) = 3^{x} + 7xy \). This function takes two variables, \( x \) and \( y \), as inputs and calculates the output by raising 3 to the power of \( x \) and adding the product of 7, \( x \), and \( y \).

Step 2: Calculate \( f(0, -2) \)

Substitute \( x = 0 \) and \( y = -2 \) into the function. \[ f(0, -2) = 3^{0} + 7 \cdot 0 \cdot (-2) = 1 + 0 = 1 \]Using the property that any number raised to the power of 0 is 1, and since the product term is multiplied by 0, it becomes zero.

Step 3: Calculate \( f(-2, 1) \)

Substitute \( x = -2 \) and \( y = 1 \) into the function. \[ f(-2, 1) = 3^{-2} + 7 \cdot (-2) \cdot 1 = \frac{1}{3^2} - 14 = \frac{1}{9} - 14 \]Since \( 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \). You calculate and simplify the expression to find the result.

Step 4: Simplify \( f(-2, 1) \)

Combine the results from the previous step. \[ rac{1}{9} - 14 = rac{1}{9} - rac{126}{9} = rac{1 - 126}{9} = rac{-125}{9} \]The expression is simplified by converting -14 to a fraction with a common denominator of 9.

Step 5: Calculate \( f(2, 1) \)

Substitute \( x = 2 \) and \( y = 1 \) into the function. \[ f(2, 1) = 3^{2} + 7 \cdot 2 \cdot 1 = 9 + 14 = 23 \]Here, \( 3^2 = 9 \) and multiplying gives \( 14 \), then add these results together.

Key Concepts

Function Evaluation

In multivariable calculus, a function like \( f(x, y) = 3^{x} + 7xy \) is evaluated by substituting specific values for its variables. This process helps in finding outputs for given inputs.

Evaluation entails replacing \( x \) and \( y \) with numbers and computing the result using the defined operations in the function.

Here's how function evaluation works:

• For \( f(0, -2) \): Replace \( x \) with 0 and \( y \) with -2 in the expression, simplifying each part.
• Always follow arithmetic operations like exponentiation and multiplication to achieve the correct output.
• Each value within the function is substituted, calculated, and combined as directed by the function's structure.

Function evaluations can reveal important properties and values, especially when used in tandem with concepts like derivatives in calculus.

Exponents and Powers

Exponents and powers are an essential part of any function evaluation in calculus, as they dictate repeated multiplication.

Understanding these concepts helps in simplifying expressions and obtaining the correct solution. Key points include:

• Any base raised to the power of zero is always 1. For example, \(3^0 = 1\).
• Negative exponents indicate a reciprocal. For example, \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}\).
• Positive powers represent standard multiplication, like \( 3^2 = 9 \).

Recognizing these exponent rules ensures solutions are consistent with algebraic principles. Strengthening these skills leads to fewer errors and more efficient problem-solving in multivariable calculus.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form, making calculations and interpretations easier. It is crucial in solving complex problems by eliminating unnecessary complexity.

Here’s how simplification plays a role in multivariable calculus function evaluation:

• Convert multiplications and additions into a combined expression, as seen in \(7xy\).
• Use common denominators for fractions—like converting \(-14\) to \(\frac{-126}{9}\) for subtraction \(\left(\frac{1}{9} - 14\right)\).
• Simplify the expression step by step to avoid mistakes, reducing to the simplest numeric result possible.

These processes not only enhance computation skills but also enable more accurate results which are foundational for deeper topics like integration and differentiation in calculus.