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For each function, evaluate (a) \(g(0,0,0)\); (b) \(g(1,0,0) ;\) (c) \(g(0,1,0) ;\) (d) \(g(z, x, y)\); (e) \(g(x+h, y+k, z+l)\), provided such a value exists. $$ g(x, y, z)=\frac{e^{x y z}}{x+y+z} $$

Short Answer

Expert verified
In summary, we have: (a) \(g(0, 0, 0)\) is undefined. (b) \(g(1, 0, 0) = 1\). (c) \(g(0, 1, 0) = 1\). (d) \(g(z, x, y) = \frac{e^{yxz}}{z+x+y}\). (e) \(g(x+h, y+k, z+l) = \frac{e^{(x+h)(y+k)(z+l)}}{(x+h)+(y+k)+(z+l)}\).

Step by step solution

01

1. Evaluate g at (0,0,0)#

To evaluate g at (0,0,0), replace x, y, and z with 0 in the expression. $$ g(0, 0, 0) = \frac{e^{(0)(0)(0)}}{(0)+(0)+(0)} $$
02

2. Simplify the expression for g(0,0,0)#

Simplify the above equation by evaluating the exponential term and the denominator. $$ g(0, 0, 0) = \frac{e^{0}}{0} $$ Since the denominator is 0, \(g(0, 0, 0)\) is undefined.
03

3. Evaluate g at (1,0,0)#

To evaluate g at (1,0,0), replace x with 1, and y and z with 0 in the expression. $$ g(1, 0, 0) = \frac{e^{(1)(0)(0)}}{(1)+(0)+(0)} $$
04

4. Simplify the expression for g(1,0,0)#

Simplify the above equation by evaluating the exponential term and the denominator. $$ g(1, 0, 0) = \frac{e^0}{1} = \frac{1}{1} = 1 $$
05

5. Evaluate g at (0,1,0)#

To evaluate g at (0,1,0), replace x and z with 0, and y with 1 in the expression. $$ g(0, 1, 0) = \frac{e^{(0)(1)(0)}}{(0)+(1)+(0)} $$
06

6. Simplify the expression for g(0,1,0)#

Simplify the above equation by evaluating the exponential term and the denominator. $$ g(0, 1, 0) = \frac{e^0}{1} = \frac{1}{1} = 1 $$
07

7. Evaluate g at (z,x,y)#

To evaluate g at (z,x,y), replace x with z, y with x, and z with y in the expression. $$ g(z, x, y) = \frac{e^{yxz}}{z+x+y} $$
08

8. Evaluate g at (x+h, y+k, z+l)#

To evaluate g at (x+h, y+k, z+l), replace x with x+h, y with y+k, and z with z+l in the expression. $$ g(x+h, y+k, z+l) = \frac{e^{(x+h)(y+k)(z+l)}}{(x+h)+(y+k)+(z+l)} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are at the heart of multivariable calculus, providing a way to measure how a function changes in relation to each variable independently. When dealing with a function like g(x, y, z) = e^{xyz} / (x+y+z), calculating the partial derivative involves taking the derivative with respect to one variable while treating the other variables as constants.

For instance, the partial derivative of g with respect to x would indicate how g's value changes as x changes, when y and z are fixed. Although the step-by-step solution does not go into deriving the partial derivatives, understanding this concept is crucial for analyzing functions in multidimensional space and optimizing them for various applications.
Exponential Functions
Exponential functions, such as e^{xyz} in our function g, are fundamental in various fields of study including mathematics, physics, and economics. These functions are characterized by a constant base (in this case, Euler's number e, approximately equal to 2.71828) raised to a power. One key feature is that the rate of growth (or decay if the exponent is negative) of exponential functions is proportional to their current value.

Moreover, exponential functions are everywhere-differentiable, meaning you can calculate their derivative at every point in their domain. As exhibited in the exercise, evaluating an exponential function where the exponent is zero yields one, hence the simplification of g(1, 0, 0) and g(0, 1, 0) results in one.
Limits of Functions
The concept of limits is a foundational one in calculus, especially when analyzing behaviors of functions as they approach certain points or infinity. In the context of the exercise, when we evaluate g(0, 0, 0), we're faced with a situation where the limit becomes critical. The function's denominator approaches zero, which hints at a potential undefined value or infinity.

Typically, when the denominator of a fraction approaches zero, we look for the limit to understand the behavior of the function. Only through limit analysis can we conclude whether the function approaches a finite value, infinity, or if it does not exist. Unfortunately, in this instance, no limit could save the function from being undefined at the origin, showcasing the pivotal role limits play in determining the continuity and behavior of functions at specific points.

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