/*! 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 33 Let \(H(x)=1-2 x^{2}\). Find the... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 33

Let \(H(x)=1-2 x^{2}\). Find the following. (a) \(H(\sqrt{2})\) (c) \(H(x+1)\) (b) \(H(5 / 6)\) (d) \(H(x+h)\)

Short Answer

Expert verified
(a) -3, (b) -7/18, (c) -2x^2 - 4x - 1, (d) -2x^2 - 4xh - 2h^2

Step by step solution

01

Evaluate H(鈭�2)

To find \(H(\sqrt{2})\), substitute \(x = \sqrt{2}\) into the function \(H(x) = 1 - 2x^2\). This gives:\[ H(\sqrt{2}) = 1 - 2(\sqrt{2})^2 = 1 - 2 \times 2 = 1 - 4 = -3. \]
02

Evaluate H(5/6)

To find \(H(\frac{5}{6})\), substitute \(x = \frac{5}{6}\) into the function \(H(x) = 1 - 2x^2\). This gives:\[ H\left(\frac{5}{6}\right) = 1 - 2\left(\frac{5}{6}\right)^2 = 1 - 2 \times \frac{25}{36} = 1 - \frac{50}{36}. \]Simplifying \(1 - \frac{50}{36}\), convert 1 to \(\frac{36}{36}\):\[ \frac{36}{36} - \frac{50}{36} = \frac{36 - 50}{36} = \frac{-14}{36} = \frac{-7}{18}. \]
03

Simplify H(x+1)

To find \(H(x+1)\), substitute \(x = x + 1\) into the function \(H(x) = 1 - 2x^2\). This gives:\[ H(x+1) = 1 - 2(x+1)^2. \]Expand \((x+1)^2\):\[ (x+1)^2 = x^2 + 2x + 1. \]Substitute back and simplify:\[ H(x+1) = 1 - 2(x^2 + 2x + 1) = 1 - 2x^2 - 4x - 2 = -2x^2 - 4x - 1. \]
04

Simplify H(x+h)

To find \(H(x+h)\), substitute \(x = x + h\) into the function \(H(x) = 1 - 2x^2\). This gives:\[ H(x+h) = 1 - 2(x+h)^2. \]Expand \((x+h)^2\):\[ (x+h)^2 = x^2 + 2xh + h^2. \]Substitute back and simplify:\[ H(x+h) = 1 - 2(x^2 + 2xh + h^2) = 1 - 2x^2 - 4xh - 2h^2. \]

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Key Concepts

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

Function Evaluation
Function evaluation is about calculating the output of a function for a given input. Let's say you have a function like \(H(x) = 1 - 2x^2\), you can determine the value of \(H\) for any specific number. For example, to evaluate \(H(\sqrt{2})\), simply replace \(x\) with \(\sqrt{2}\) in the function.
  • Start with the original function: \(H(x) = 1 - 2x^2\).
  • Substitute \(x = \sqrt{2}\): \(H(\sqrt{2}) = 1 - 2(\sqrt{2})^2\).
  • Simplify: \(H(\sqrt{2}) = 1 - 2 \times 2 = 1 - 4 = -3\).
This shows the value of the function for \(x = \sqrt{2}\) is -3. Function evaluation helps us understand the behavior of functions with different inputs.
Substitution in Functions
Substitution in functions is a method used to replace a variable in a function with a specific value or another expression. When dealing with functions, this process helps to determine how changes in the input affect the output. For instance, to evaluate \(H(5/6)\), which involves substituting \(x\) with \(\frac{5}{6}\) in \(H(x) = 1 - 2x^2\).
  • Begin with the function: \(H(x) = 1 - 2x^2\).
  • Replace \(x\) with \(\frac{5}{6}\): \(H\left(\frac{5}{6}\right) = 1 - 2\left(\frac{5}{6}\right)^2\).
  • Calculate: \(H\left(\frac{5}{6}\right) = 1 - 2 \times \frac{25}{36}\).
  • Simplify further: \(= 1 - \frac{50}{36} = \frac{36}{36} - \frac{50}{36} = \frac{-14}{36} = \frac{-7}{18}\).
Thus, the substitution process gives \(H\left(\frac{5}{6}\right) = \frac{-7}{18}\) as the output.
Quadratic Expansion
Quadratic expansion is used when you expand an expression like \((x + a)^2\). This concept is handy when evaluating functions with polynomial expressions. For example, if you want to find \(H(x+1)\) or \(H(x+h)\), you need to expand \((x+1)^2\) or \((x+h)^2\) within the function.Consider \(H(x+1) = 1 - 2(x+1)^2\).
  • First, expand: \((x+1)^2 = x^2 + 2x + 1\).
  • Now substitute into the function: \(H(x+1) = 1 - 2(x^2 + 2x + 1)\).
  • Simplify: \(= 1 - 2x^2 - 4x - 2 = -2x^2 - 4x - 1\).
Similarly, for \(H(x+h)\), after expanding \((x+h)^2\) to \(x^2 + 2xh + h^2\):
  • Substitute: \(H(x+h) = 1 - 2(x^2 + 2xh + h^2)\).
  • Simplify: \(= 1 - 2x^2 - 4xh - 2h^2\).
Quadratic expansion helps in achieving simpler forms of more complex expressions.

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Most popular questions from this chapter

Sketch the graph of the function, given that \(f,\) \(\mathrm{F}\) and \(\mathrm{g}\) are defined as follows. (Hint: Start with the basic graphs in Figure 7 on page 149.) $$f(x)=|x| \quad F(x)=1 / x \quad g(x)=\sqrt{1-x^{2}}$$ $$y=1-g(x-2)$$

Let \(f(x)=-\frac{2 x+2}{x}\) (a) Find \(f[f(x)]\) (b) Use a graphing utility to graph \(y=f[f(x)] .\) Display the graph using true proportions. What type of symmetry does the graph appear to have? (c) The result in part (b) suggests that the inverse of the function \(f \circ f\) is again \(f \circ f .\) Use algebra to show that this is indeed correct.

Let \(f(x)=-2 x+1\) and \(g(x)=a x+b .\) Find \(a\) and \(b\) so the equation \(f[g(x)]=x\) holds for all values of \(x\).

$$\text { Let } f(z)=\frac{3 z-4}{5 z-3} . \text { Find } f\left(\frac{3 z-4}{5 z-3}\right)$$

\(\pi=3.141592653589793 \ldots\) and so on! For each natural number \(n,\) let \(G(n)\) be the digit in the \(n\) th decimal place of \(\pi .\) For instance, according to the expression for \(\pi\) given above, we have \(G(1)=1, G(2)=4,\) and \(G(5)=9\) (a) Use the expression for \(\pi\) given above to evaluate \(G(10)\) and \(G(14)\) (b) Use the Internet to help you evaluate \(G(100), G(750)\) and \(G(1000) .\) Suggestions: Using any of the common search engines on the World Wide Web, under the categories of mathematics or science, search for "pi." Here, for example, are two sites that contain the information you need. (They were accessible at the time of this writing, March 2004 .)
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