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Chapter 2: Problem 3

Find \(f+g, f-g, f g,\) and \(f / g\) and their domains. $$f(x)=\sqrt{4-x^{2}}, \quad g(x)=\sqrt{1+x}$$

Short Answer

Expert verified
Domains: \(f+g, f-g, fg: [-1, 2]; f/g: (-1, 2]\).

Step by step solution

01

Identify the Functions

First, identify the given functions: \(f(x) = \sqrt{4-x^2}\) and \(g(x) = \sqrt{1+x}\).
02

Find the Domain of f(x)

The function \(f(x) = \sqrt{4-x^2}\) is defined where the expression under the square root is non-negative. Set the inequality:\[4 - x^2 \geq 0\] which implies \[-2 \leq x \leq 2\]. Thus, the domain of \(f(x)\) is \([-2, 2]\).
03

Find the Domain of g(x)

For \(g(x) = \sqrt{1+x}\), the expression under the square root needs to be non-negative:\[1 + x \geq 0\] which implies \(x \geq -1\). Therefore, the domain of \(g(x)\) is \([-1, \infty)\).
04

Find f+g and Its Domain

The sum \((f+g)(x) = \sqrt{4-x^2} + \sqrt{1+x}\) is defined where both \(f(x)\) and \(g(x)\) are defined. This is the intersection of their domains: \([-2, 2]\) and \([-1, \infty)\). The intersection is \([-1, 2]\).
05

Find f-g and Its Domain

The subtraction \((f-g)(x) = \sqrt{4-x^2} - \sqrt{1+x}\) also requires the intersection of the domains of \(f(x)\) and \(g(x)\), which is \([-1, 2]\).
06

Find fg and Its Domain

The product \((fg)(x) = \sqrt{4-x^2} \cdot \sqrt{1+x}\) is defined over the same domain as \(f+g\) and \(f-g\), which is \([-1, 2]\).
07

Find f/g and Its Domain

For the division \(\left(\frac{f}{g}\right)(x) = \frac{\sqrt{4-x^2}}{\sqrt{1+x}}\), \(g(x) eq 0\). This occurs when \(1+x > 0\), or \(x > -1\). Therefore, the domain for \(\frac{f}{g}\) is \((-1, 2]\).

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Key Concepts

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

Understanding Function Addition
Function addition involves combining two functions into one by adding their outputs. Let's say we have two functions, \(f(x)\) and \(g(x)\). To find the function \((f+g)(x)\), we evaluate both functions at the same input \(x\) and add the results:
  • \((f+g)(x) = f(x) + g(x)\)
In our exercise, \(f(x) = \sqrt{4-x^2}\) and \(g(x) = \sqrt{1+x}\), so:
  • \((f+g)(x) = \sqrt{4-x^2} + \sqrt{1+x}\)
The domain of the sum \((f+g)\) is where both \(f(x)\) and \(g(x)\) are defined. This means we need the values of \(x\) that satisfy both domains. For \(f(x)\), \(-2 \leq x \leq 2\), and for \(g(x)\), \(x \geq -1\). The intersection, where both conditions hold true, is \([-1, 2]\). This is the domain for \((f+g)(x)\).
Understanding Function Subtraction
Function subtraction is quite similar to function addition, but instead of adding, we subtract the outputs of the functions:
  • \((f-g)(x) = f(x) - g(x)\)
Using our functions, we write:
  • \((f-g)(x) = \sqrt{4-x^2} - \sqrt{1+x}\)
The domain for function subtraction is the set of \(x\) values where both functions are defined, which again is the intersection of their domains. For \(f(x)\) and \(g(x)\), this range is from \(-1\) to \(2\), i.e., \([-1, 2]\). This means \(f-g\) can only be evaluated within this interval.
Understanding Function Multiplication
When multiplying functions, each input is used to find the product of their outputs:
  • \((fg)(x) = f(x) \cdot g(x)\)
For our provided functions, the multiplication is:
  • \((fg)(x) = \sqrt{4-x^2} \cdot \sqrt{1+x}\)
Function multiplication also requires both functions to be defined over the same domain. Hence, we take the intersection of their individual domains. For \(f(x)\) and \(g(x)\), this is \([-1, 2]\). Therefore, \((fg)(x)\) is valid within this range.
Understanding Function Division
Function division involves dividing the outputs of two functions:
  • \((f/g)(x) = \frac{f(x)}{g(x)}\)
Using our functions, we have:
  • \(\left(\frac{f}{g}\right)(x) = \frac{\sqrt{4-x^2}}{\sqrt{1+x}}\)
Division requires that \(g(x) eq 0\). Thus, we look for values of \(x\) where \(\sqrt{1+x} > 0\), which simplifies to \(x > -1\). Considering the intersection of domains \([-2, 2]\) for \(f(x)\) and \((-1, \infty)\) for \(g(x) > 0\), the domain of the division is \((-1, 2]\). This means division can be safely performed where \(x > -1\) and \(\leq 2\).

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Most popular questions from this chapter

In a certain state the maximum speed permitted on freeways is \(65 \mathrm{mi} / \mathrm{h}\) and the minimum is \(40 .\) The fine \(F\) for violating these limits is \(\$ 15\) for every mile above the maximum or below the minimum. (a) Complete the expressions in the following piecewise defined function, where \(x\) is the speed at which you are driving. $$F(x)=\left\\{\begin{array}{ll}\text {\(\text{________ }\) if } 0 < x < 40 \\\\\text {\(\text{________ }\) if } 40 \leq x \leq 65 \\\\\text {\(\text{________ }\) if } x > 65\end{array}\right.$$ (b) Find \(F(30), F(50),\) and \(F(75)\). (c) What do your answers in part (b) represent?

Find a function whose graph is the given curve. The bottom half of the circle \(x^{2}+y^{2}=9\)

Volume of a Box A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side \(x\) at each comer and then folding up the sides (see the figure). (a) Find a function that models the volume of the box. (b) Find the values of \(x\) for which the volume is greater than 200 in \(^{3}\). (c) Find the largest volume that such a box can have.

In the margin notes in this section we pointed out that the inverse of a function can be found by simply reversing the operations that make up the function. For instance, in Example 6 we saw that the inverse of $$f(x)=3 x-2 \quad \text { is } \quad f^{-1}(x)=\frac{x+2}{3}$$ because the "reverse" of "multiply by 3 and subtract 2" is "add 2 and divide by 3 " Use the same procedure to find the inverse of the following functions. (a) \(f(x)=\frac{2 x+1}{5}\) (b) \(f(x)=3-\frac{1}{x}\) (c) \(f(x)=\sqrt{x^{3}+2}\) (d) \(f(x)=(2 x-5)^{3}\) Now consider another function: \(f(x)=x^{3}+2 x+6\) Is it possible to use the same sort of simple reversal of operations to find the inverse of this function? If so, do it. If not, explain what is different about this function that makes this task difficult.

Express the function in the form \(f \circ g\) $$G(x)=\frac{x^{2}}{x^{2}+4}$$
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