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Chapter 10: Problem 11

Evaluate each function at the given point. \(f(x, y)=\sqrt{2 x+3 y^{2}}\) at \((-1,2)\)

Short Answer

Expert verified
The function value at (-1, 2) is \( \sqrt{10} \).

Step by step solution

01

Identify the Function and Point

We are given the function \[ f(x, y) = \sqrt{2x + 3y^2} \]and we need to evaluate it at the given point \[ (-1, 2) \].
02

Substitute the Values

Substitute \(x = -1\) and \(y = 2\) into the function:\[ f(-1, 2) = \sqrt{2(-1) + 3(2)^2} \]
03

Calculate Inside the Square Root

Calculate the expression inside the square root:\[ 2(-1) + 3(2)^2 = -2 + 3(4) = -2 + 12 = 10 \]
04

Compute the Square Root

Finally, compute the square root of 10:\[ f(-1, 2) = \sqrt{10} \]

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Key Concepts

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

Function Evaluation
Function evaluation allows us to find out what a function equals at a specific point. In other words, we plug in given values for the variables to see what output we get from the function. Here鈥檚 how it generally works:
  • You are given a function, for example, \(f(x, y) = \sqrt{2x + 3y^2}\).
  • You also have a specific point at which you need to evaluate the function, such as \((-1, 2)\).
  • Substitute the given values into the function to replace the variables.
This process helps understand how the function behaves with different inputs. By seeing the result, you comprehend the value of the function at that given point. This is a fundamental skill in calculus that aids in understanding more complex problems involving limits and optimization.
Square Root Function
The square root function is a common mathematical operation. It involves finding a number that when multiplied by itself gives the original number inside the square root. In our function \(f(x, y) = \sqrt{2x + 3y^2}\), we encounter a square root expression.
  • The expression inside the square root is evaluated first.
  • Once evaluated, you find the square root of that result.
  • This gives you the final value of the function at the specified point.
The square root function is an essential part of calculus and algebra. It helps with equations involving powers and roots. Understanding how to compute square roots is crucial whether dealing with simple operations or more complex calculus problems.
Substitution Method
The substitution method is used to simplify the process of evaluating functions. It involves replacing variables with given values to calculate the result. Let's break it down:
  • Identify the function and the point you need to evaluate, such as \(f(x, y) = \sqrt{2x + 3y^2}\) and \((-1, 2)\).
  • Substitute \(x = -1\) and \(y = 2\) into the function. Insert each value directly where the matching variable appears.
  • Calculate the resulting expression step by step, until the function is simplified enough to evaluate fully.
This method simplifies the evaluation of functions by breaking the process into manageable steps. Substitution is widely used in calculus for solving equations, working on integrals, and even transforming complex expressions into simpler forms.

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Most popular questions from this chapter

At the beginning of this chapter we introduced the heat index as a way of calculating how temperature and humidity affect the apparent temperature. The equation for the heat index is: \(\begin{aligned} H(T, R)=&-42.38+2.049 T+10.14 R-6.838 \times 10^{-3} T^{2} \\\ &-0.2248 T R-5.482 \times 10^{-2} R^{2}+1.229 \times 10^{-3} T^{2} R \\\ &+8.528 \times 10^{-4} T R^{2}-1.99 \times 10^{-6} T^{2} R^{2} \end{aligned}\) where \(T\) is the actual air temperature (in \({ }^{\circ} \mathrm{F}\) ) and \(R\) is the relative humidity (in \%). Using nine evenly spaced points and five colors, make a heat map for the heat index for the domain \(D=\\{(T, R):\) \(80 \leq T \leq 100,40 \leq R \leq 60]\). (You will find it easiest to calculate the heat index, \(H\), if you program the formula for the heat index into a graphing calculator.)

Find the indicated partial derivatives. \(f(x, y)=\sin (3 x y) ; \frac{\partial^{2} f}{\partial y^{2}}\)

Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(-1.1)}\left(2 x y+y^{2}\right)\)

Find the range of each function \(f(x, y)\), when defined on the specified domain \(D\). \(f(x, y)=x-y^{2} ; D=\\{(x, y):-1 \leq x \leq 1,1 \leq y \leq 2\\}\)

Find the range of each function \(f(x, y)\), when defined on the specified domain \(D\). \(f(x, y)=\frac{x}{y} ; D=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq 1, y>x]\)
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