Problem 9 - Calculus Early Transcendentals

Question

Let $$f(x, y, z)=e^{\sqrt{1-x^{2}-y^{2}}}$$ (a) Evaluate \(f(2,-1,6), \quad\) (b) Find the domain of \(f\) (c) Find the range of \(f\)

Step-by-Step Solution

Answer

(a) Undefined, (b) Domain: \( x^{2} + y^{2} \leq 1 \), (c) Range: \([1, e]\).

Step 1: Evaluate f(2, -1, 6)

To evaluate \( f(2, -1, 6) \), substitute \( x = 2 \), \( y = -1 \), and \( z = 6 \) into the function \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \). This gives:\[ f(2, -1, 6) = e^{\sqrt{1-(2)^{2}-(-1)^{2}}} = e^{\sqrt{1-4-1}} = e^{\sqrt{-4}} \]Since \( \sqrt{-4} \) is not a real number, \( f(2, -1, 6) \) is undefined for real numbers.

Step 2: Identify the Domain of f

The domain of \( f \) consists of all points \( (x, y, z) \) such that the expression under the square root in \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \) is non-negative, i.e., \( 1-x^{2}-y^{2} \geq 0 \). Simplifying, we find:\[ x^{2} + y^{2} \leq 1 \]Thus, the domain of \( f \) is all triples \( (x, y, z) \) where \( x^{2} + y^{2} \leq 1 \), for any real \( z \).

Step 3: Determine the Range of f

Since \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \) and \( \sqrt{1-x^{2}-y^{2}} \) can range from 0 to 1 (because \( x^{2} + y^{2} \leq 1 \)), calculate \( e \) raised to these values. \( e^{0} = 1 \) and \( e^{1} = e \). Therefore, the range of \( f \) is the interval \( [1, e] \).

Key Concepts

Function Evaluation

Function evaluation is all about substituting the values you are given into a function to see what result you get. To evaluate a function like \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \), you simply replace each variable with the given numbers. For example, if we want to evaluate \( f(2, -1, 6) \), we substitute \( x = 2 \), \( y = -1 \), and \( z = 6 \). After doing this, our function becomes \( e^{\sqrt{1-2^2-(-1)^2}} \). This simplifies to \( e^{\sqrt{1-4-1}} = e^{\sqrt{-4}} \).

However, there's a catch! The square root of a negative number is not real. This tells us that \( f(2, -1, 6) \) doesn't give us a real-world number. In this case, the function is undefined for these inputs, which is a crucial insight.

Remember: Always take note of whether the value you are substituting will lead to a valid computation. If you can't, the function at those values isn't defined.

Domain of a Function

The domain of a function represents all the possible input values (or solutions) where the function works properly. For the function \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \), the main concern lies within the square root in the formula. Remember, the expression under a square root must be non-negative if we're considering real numbers.

To determine the domain:

• Look at \( 1-x^{2}-y^{2} \) and set it \( \geq 0 \).
• This simplifies to \( x^{2} + y^{2} \leq 1 \).

So, any triple \( (x, y, z) \) where \( x^2 + y^2 \leq 1 \) is in the domain. And here's the neat bit: the third variable, \( z \), doesn't even affect the function, so it can be any real number.

The domain is a circular region for \( x \) and \( y \) where they stay within this boundary, and \( z \) can be any real number.

Range of a Function

To find the range of a function means to determine all possible output values created by the function. For our function, \( f(x, y, z) = e^{\sqrt{1-x^{2}-y^{2}}} \), the range is derived from what \( \sqrt{1-x^{2}-y^{2}} \) can output.

Based on the domain \( x^{2} + y^{2} \leq 1 \), the expression inside the square root can vary from 0 (at the edges of the domain) to 1 (at the center where \( x = 0 \) and \( y = 0 \)).

Using these limits:

• If \( \sqrt{1-x^{2}-y^{2}} = 0 \), then \( e^0 = 1 \).
• If \( \sqrt{1-x^{2}-y^{2}} = 1 \), then \( e^1 = e \).

This tells us the function's output (or range) is between 1 and \( e \), including both. Hence, the range of \( f \) is \([1, e]\). This means for any input within the domain, the result of the function will always be somewhere in or on the edges of this range.