Problem 16
Question
Evaluate each function at the given point. \(g\left(x_{1}, x_{2}, x_{3}\right)=x_{1} \sqrt{x_{2} x_{3}}\) at \((1,2,1)\)
Step-by-Step Solution
Verified Answer
The function evaluates to \(\sqrt{2}\).
1Step 1: Substitute Variables
We start by substituting the given values of the variables into the function. Here, substitute \(x_1 = 1\), \(x_2 = 2\), and \(x_3 = 1\) into the function \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\).
2Step 2: Calculate the Product Inside the Square Root
Calculate the product of \(x_2\) and \(x_3\). This is \(2 \times 1 = 2\). Hence, the expression inside the square root becomes \(\sqrt{2}\).
3Step 3: Evaluate the Function
Multiply \(x_1\) by the square root expression. This results in \(1 \times \sqrt{2}\), which simplifies to \(\sqrt{2}\).
Key Concepts
Function EvaluationSubstitution MethodSquare Root Calculation
Function Evaluation
When we talk about evaluating a function, we mean finding the output for given input values. In multivariable calculus, functions often have more than one input variable. The function provided, \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\), is a function of three variables, \(x_1\), \(x_2\), and \(x_3\).
To evaluate the function, you need to substitute the specific values of \(x_1\), \(x_2\), and \(x_3\) as given. This process might seem straightforward, but it is crucial because missing or misplacing values would lead to incorrect outcomes.
To evaluate the function, you need to substitute the specific values of \(x_1\), \(x_2\), and \(x_3\) as given. This process might seem straightforward, but it is crucial because missing or misplacing values would lead to incorrect outcomes.
- Identify all the variables in the given function.
- Plug in the numbers provided for each respective variable.
- Follow mathematical operations as per the function's definition.
Substitution Method
The substitution method is a staple technique in evaluating functions, especially in calculus. It involves replacing each variable within a function with its corresponding numerical value.
For the function \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\), the given point is \((1, 2, 1)\). This tells us:
1. Replace \(x_1\) in the function with 1.
2. Replace \(x_2\) with 2 and \(x_3\) with 1.
3. This turns \(g(x_1, x_2, x_3)\) into an expression you can solve: \(1 \times \sqrt{2 \times 1}\).
Substitution is not just about inserting values randomly. It maintains the function's integrity while personalizing it for specific data points, making this method invaluable in problem-solving scenarios.
For the function \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\), the given point is \((1, 2, 1)\). This tells us:
- \(x_1 = 1\)
- \(x_2 = 2\)
- \(x_3 = 1\)
1. Replace \(x_1\) in the function with 1.
2. Replace \(x_2\) with 2 and \(x_3\) with 1.
3. This turns \(g(x_1, x_2, x_3)\) into an expression you can solve: \(1 \times \sqrt{2 \times 1}\).
Substitution is not just about inserting values randomly. It maintains the function's integrity while personalizing it for specific data points, making this method invaluable in problem-solving scenarios.
Square Root Calculation
The square root calculation is an essential mathematical operation, often seen in multivariable functions. It involves finding a number that, when multiplied by itself, gives the original number.
In the function \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\), after substituting \(x_2 = 2\) and \(x_3 = 1\), you find the expression inside the square root as \(2 \times 1 = 2\).
Now, you have the task of evaluating \(\sqrt{2}\). This expression is the square root of 2. Although \(\sqrt{2}\) does not yield a whole number, it's approximately 1.414. However, in most mathematical contexts, retaining it as \(\sqrt{2}\) is preferable for precision.
In the function \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\), after substituting \(x_2 = 2\) and \(x_3 = 1\), you find the expression inside the square root as \(2 \times 1 = 2\).
Now, you have the task of evaluating \(\sqrt{2}\). This expression is the square root of 2. Although \(\sqrt{2}\) does not yield a whole number, it's approximately 1.414. However, in most mathematical contexts, retaining it as \(\sqrt{2}\) is preferable for precision.
- Calculate the product inside the square root.
- Find the square root of the calculated product.
- Use the result in further calculations.
Other exercises in this chapter
Problem 16
Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\ln \left(\frac{x y}{x^{2}+x y}\right)\)
View solution Problem 16
Show that $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x-y)(x+2 y)}{x^{2}+y^{2}}$$ does not exist by computing the limit along the positive \(x\) -axis and the posi
View solution Problem 16
Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\left(x^{2}+y^{2}\right) e^{-x^{2}-y^{2}} ;(1,1)\)
View solution Problem 16
The functions are defined on the rectangular domain $$D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\}$$ Find the global maxima and minima of \(f\) on \(D .\) $
View solution