/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Evaluate each function at the gi... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each function at the given point. \(g\left(x_{1}, x_{2}, x_{3}\right)=x_{1} \sqrt{x_{2} x_{3}}\) at \((1,2,1)\)

Short Answer

Expert verified
The function evaluates to \(\sqrt{2}\).

Step by step solution

01

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 \(g(x_1, x_2, x_3) = x_1 \sqrt{x_2 x_3}\).
02

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 \(\sqrt{2}\).
03

Evaluate the Function

Multiply \(x_1\) by the square root expression. This results in \(1 \times \sqrt{2}\), which simplifies to \(\sqrt{2}\).

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Key Concepts

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

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.
  • 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.
With proper substitution, you transform an abstract formula into a computable expression, making the complex output concrete and understandable.
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:
  • \(x_1 = 1\)
  • \(x_2 = 2\)
  • \(x_3 = 1\)
Using these values, you substitute them directly into the function:

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.
  • Calculate the product inside the square root.
  • Find the square root of the calculated product.
  • Use the result in further calculations.
Always remember: the square root maintains part of the mathematical symmetry that is crucial in many calculus problems, offering exactness and clarity in your computations.

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Most popular questions from this chapter

(a) Write $$h(x, y)=\sqrt{x+y}$$ as a composition of two functions.

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(-1,3)} x^{2}\left(y^{2}-3 x y\right)\)

A chemical diffuses in a container that occupies the interval \(0 \leq x \leq 1\). The concentration of the chemical at time \(t\) and at a point \(x\) is given by the diffusion equation: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ (a) Suppose that the chemical is allowed to diffuse through the entire container until the concentration reaches an equilibrium value where \(c\) does not change any more with time, that is, \(\partial c / \partial t=0 .\) Suppose that chemical that touches the walls of the container is removed so that $$ c(0, t)=c(1, t)=0 . $$ The steady state concentration of chemical will be a function \(C(x)\) with $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ and \(C(0)=C(1)=0\). Show that \(C(x)=0\) satisfies this differential equation and the constraints as the points \(x=0\) and \(x=1\). (b) Now suppose that chemical is added to the container by a reaction that occurs at the wall \(x=0 .\) This reaction keeps the concentration of chemical at this wall equal to \(c(0, t)=1 . \mathrm{Un}\) der these conditions the steady state distribution of chemical will obey a differential equation: $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ with \(C(0)=1\) and \(C(1)=0 .\) Show that \(C(x)=1-x\) satisfies both the differential equation and the boundary conditions at \(x=0\) and \(x=1\). (c) Notice that the steady state distributions in (a) and (b) do not depend on \(D .\) Can you explain why?

Compute $$\lim _{(x-y) \rightarrow(0,0)} \frac{4 x y}{x^{2}+y^{2}}$$ along the \(x\) -axis, the \(y\) -axis, and the line \(y=x\). What can you conclude?

Compute $$\lim _{(x, y) \rightarrow(0,0)} \frac{3 x(y+x)}{x^{2}+y^{3}}$$ along lines of the form \(y=m x\), for \(m \neq 0 .\) What can you conclude?

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