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Chapter 0: Problem 28

If $$ f(x)=\left\\{\begin{array}{ll} 2+\sqrt{1-x} & \text { if } x \leq 1 \\ 2 \cos 2 \pi x & \text { if } x>1 \end{array}\right. $$ find \(f(0), f(1)\), and \(f(2)\).

Short Answer

Expert verified
The short versions of the answers are: \(f(0) = 3\), \(f(1) = 2\), and \(\overline{f(2)} = 2\).

Step by step solution

01

f(0) = 2 + \(\sqrt{1-0}\) = 2 + \(\sqrt{1}\) = 2 + 1 = 3. #Step 2: Find f(1)# For x = 1, we look at the conditions for each part of the piecewise function. Since 1 鈮� 1, we will again use the first part of the function \(f(x) = 2 + \sqrt{1-x}\). Plug in x = 1:

f(1) = 2 + \(\sqrt{1-1}\) = 2 + \(\sqrt{0}\) = 2 + 0 = 2. #Step 3: Find \(\overline{f(2)}\)# For x = 2, we look at the conditions for each part of the piecewise function. Since 2 > 1, we will use the second part of the function \(f(x) = 2\cos(2\pi x)\). Plug in x = 2:
02

\(f(2) = 2\cos(2\pi (2)) = 2\cos(4\pi)\) Since the cosine function has a period of \(2\pi\) (#meaning cos(x) = cos(x+2k\(\pi\), where k is an integer#),

\(\cos(4\pi) = \cos(0)\) Finally,
03

\(f(2) = 2\cos(0) = 2(1) = 2\) Now, we need to find the conjugate of f(2): \(\overline{f(2)}\) Since f(2) is a real number (2), its conjugate is the same as itself. Thus,

\(\overline{f(2)} = 2\) So, the results are: 1. f(0) = 3 2. f(1) = 2 3. \(\overline{f(2)}\) = 2

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Most popular questions from this chapter

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ f(x)=\frac{1+x}{1-x} ; \quad g(x)=\frac{x-1}{x+1} $$

Plot the graph of the function \(f\) in (a) the standard viewing window and (b) the indicated window. $$ f(x)=x^{3}-20 x^{2}+8 x-10 ; \quad[-20,20] \times[-1200,100] $$

Find \(f^{-1}(a)\) for the function \(f\) and the real number \(a\). $$ f(x)=\frac{3}{\pi} x+\sin x ; \quad-\frac{\pi}{2}

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. \(y=\tan x, \quad y=\tan \left(x+\frac{\pi}{3}\right)\)

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. \(y=\sqrt{x}, \quad y=2 \sqrt{x-1}+1\)
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