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Chapter 3: Problem 6

Compute each expression, given that the functions f, g, h, k, and m are defined as follows: $$\begin{array}{ll}f(x)=2 x-1 & k(x)=2, \text { for all } x \\ g(x)=x^{2}-3 x-6 & m(x)=x^{2}-9 \\ h(x)=x^{3}\end{array}$$ (a) \((g+m)(x)\) (b) \((g+m)(x)-(g-m)(x)\)

Short Answer

Expert verified
\((g+m)(x) = 2x^2 - 3x - 15\) and \((g+m)(x) - (g-m)(x) = 2x^2 - 18\).

Step by step solution

01

Understand the Functions

We are given the functions: \( f(x)=2x-1 \), \( g(x)=x^2-3x-6 \), \( h(x)=x^3 \), \( k(x)=2 \), and \( m(x)=x^2-9 \). Our task is to work with the functions \( g \) and \( m \).
02

Compute \((g+m)(x)\)

The expression \((g+m)(x)\) means that we add \( g(x) \) and \( m(x) \). Therefore,\[(g+m)(x) = g(x) + m(x) = (x^2 - 3x - 6) + (x^2 - 9)\]Combine like terms:\[(g+m)(x) = x^2 + x^2 - 3x - 6 - 9 = 2x^2 - 3x - 15\]
03

Compute \((g-m)(x)\)

The expression \((g-m)(x)\) means that we subtract \( m(x) \) from \( g(x) \). Therefore,\[(g-m)(x) = g(x) - m(x) = (x^2 - 3x - 6) - (x^2 - 9)\]Distribute the subtraction:\[(g-m)(x) = x^2 - 3x - 6 - x^2 + 9 = -3x + 3\]
04

Compute \((g+m)(x) - (g-m)(x)\)

Now, we subtract \((g-m)(x)\) from \((g+m)(x)\):\[(g+m)(x) - (g-m)(x) = (2x^2 - 3x - 15) - (-3x + 3)\]Distribute the subtraction:\[2x^2 - 3x - 15 + 3x - 3 = 2x^2 - 18\]
05

Final Answer

Thus, we have found both parts of the problem. \((g+m)(x) = 2x^2 - 3x - 15\) and \((g+m)(x) - (g-m)(x) = 2x^2 - 18\).

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Key Concepts

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

Addition of Polynomials
When you add polynomials, you are essentially combining like terms. Think of each polynomial as a list of items that you are grouping together. In our exercise, we need to add the polynomials \(g(x)\) and \(m(x)\).
\(g(x) = x^2 - 3x - 6\) and \(m(x) = x^2 - 9\). To add them, align the like terms:
  • The \(x^2\) terms: \( x^2 + x^2 = 2x^2 \)
  • The \(x\) terms: There is only \(-3x\) from \(g(x)\) as \(m(x)\) has no \(x\) term.
  • The constant terms: \(-6 + (-9) = -15\)
So, after adding, we get \((g+m)(x) = 2x^2 - 3x - 15\). Adding polynomials doesn鈥檛 require you to rearrange terms, just line them up and combine.
Subtraction of Polynomials
Subtracting polynomials involves changing signs and then combining like terms. It's similar to addition but with a few more steps. Let's subtract \(m(x)\) from \(g(x)\) in our example.
We have \(g(x) = x^2 - 3x - 6\) and \(m(x) = x^2 - 9\). To find \((g - m)(x)\), subtract each corresponding term:
  • The \(x^2\) terms are subtracted: \(x^2 - x^2 = 0\)
  • The \(x\) terms remain: \(-3x\)
  • For the constants, subtract: \(-6 - (-9) = -6 + 9 = 3\)
Therefore, \((g-m)(x) = -3x + 3\). Remember, subtracting polynomials is like distributing a negative sign and then combining terms.
Function Operations
Function operations involve performing arithmetic with functions similar to how we do with numbers, like adding or subtracting. In our case, we combined functions \(g(x)\) and \(m(x)\) using addition and subtraction.
We computed:
  • \((g+m)(x)\) as described in the addition section, which gave us \(2x^2 - 3x - 15\).
  • \((g-m)(x)\), explained in the subtraction section, resulted in \(-3x + 3\).
To find \((g+m)(x) - (g-m)(x)\), we subtract \((g-m)(x)\) from \((g+m)(x)\):
  • Put \(-3x + 3\) in parentheses to appreciate the change of signs when subtracting.
  • The expression becomes \((2x^2 - 3x - 15) - (-3x + 3)\), which simplifies to \(2x^2 + 3x - 3 - 15\).
  • After calculating, you get \(2x^2 - 18\).
Function operations allow you to view functions as individual entities that can participate in algebraic expressions, enabling you to manipulate them easily.

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Most popular questions from this chapter

(a) Find the difference quotient \(\frac{f(x)-f(a)}{x-a}\) for each function, as in Example 4. (b) Find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each function, as in Example \(5 .\) $$f(x)=-3 / x^{2}$$

Let \(q(x)=a x^{2}+b x+c .\) Evaluate $$q\left(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\right)$$

$$\text { If } f(x)=-2 x^{2}+6 x+k \text { and } f(0)=-1, \text { find } k$$

Suppose that \((a, b)\) is a point on the graph of \(y=f(x)\) Match the functions defined in the left-hand column with the points in the right-hand column. For example, the appropriate match for (a) in the left-hand column is determined as follows. The graph of \(y=f(x)+1\) is obtained by translating the graph of \(y=f(x)\) up one unit. Thus the point \((a, b)\) moves up to \((a, b+1)\), and consequently, (E) is the appropriate match for (a). (a) \(y=f(x)+1 \quad\) (A) \((-a, b)\) (b) \(y=f(x+1)\) (B) \((b, a)\) (c) \(y=f(x-1)+1\) (C) \((a-1, b)\) (d) \(y=f(-x)\) (D) \((-b, a+1)\) (e) \(y=-f(x)\) (E) \((a, b+1)\) (f) \(y=-f(-x)\) (F) \((1-a, b)\) (g) \(y=f^{-1}(x)\) (G) \((-a,-b)\) (h) \(y=f^{-1}(x)+1\) (H) \((-b, 1-a)\) (i) \(y=f^{-1}(x-1)\) (I) \((b,-a)\) (j) \(y=f^{-1}(-x)+1\) (J) \((a,-b)\) (k) \(y=-f^{-1}(x)\) (K) \((b+1, a)\) (1) \(\quad y=-f^{-1}(-x)+1\) (L) \((a+1, b+1)\) (m) \(y=1-f^{-1}(x)\) (M) \((b, a+1)\) (n) \(y=f(1-x)\) (N) \((b, 1-a)\)

Indicate how iteration is used in finding roots of numbers and roots of equations. (The functions that are given in each exercise were determined using Newton's method, a process studied in calculus.) Let \(f(x)=0.5\left(x+\frac{3}{x}\right)\). (a) Compute the first ten iterates of \(x_{0}=1\) under the function \(f .\) What do you observe? (b) Use your calculator to evaluate \(\sqrt{3}\) and compare the answer to your results in part (a). What do you observe? (c) It can be shown that for any positive number \(x_{0}\), the iterates of \(x_{0}\) under the function \(f(x)=0.5(x+3 / x)\) always approach the number \(\sqrt{3}\). (You'll see the reasons for this in Section 4.3.) Looking at your results in parts (a) and (b), which is the first iterate that agrees with \(\sqrt{3}\) through the first three decimal places? Through the first eight decimal places? (d) Compute the first ten iterates of \(x_{0}=50\) under the function \(f\), then answer the questions presented in part (c).
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