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Sketch the following graphs using the graph of \(f(x)=x^3\):
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\[ y = x^3 - 3 \]
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\[ y = (x -1)^3 \]
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\[ y = -x^3 + 1 \]
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\[ y = -(x-1)^3 + 1 \]
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Sketch the following graphs using the graph of \(f(x)=\cfrac{1}{x}\):
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\[ y = \frac{1}{x} -2 \]
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\[ y = \frac{1}{x-2} \]
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\[ y = - \frac{1}{x} \]
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\[ y = \frac{1}{x-2} +5 \]
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Sketch the following graphs using the graph of \( y = \lfloor x \rfloor \):
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\[ y = - \lfloor x \rfloor \]
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\[ y = \lfloor 2x \rfloor \]
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\[ y = \frac{1}{2} \lfloor x \rfloor \]
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Using the translation and reflection techniques, and the graph of \(y=\sin x\), graph the function \(y=1- \sin(x - \frac{\pi}{2})\).
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Considering the graph of \(y= \cos x\):
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sketch the graph of \(y = -3\cos 4x\) using transformation techniques.
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find the period of \(y = -3\cos 4x\).
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In the exercises, from 6 to 8, find \(\boldsymbol{f+g,\; f-g,\; fg}\) and \(\boldsymbol{f/g}\), with their respective domains.
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\[ f(x) = \cfrac{1}{1-x},\quad g(x) = \sqrt{2-x} \]
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\[ f(x) = \sqrt{16-x^2},\quad g(x) = \sqrt{x^2-4} \]\[ \begin{aligned} &f(x) = \sqrt{16-x^2}, \\[1em] &g(x) = \sqrt{x^2-4} \end{aligned} \]
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\[ f(x) = \cfrac{1}{\sqrt{4-x^2}},\quad g(x) = \sqrt[3]{x} \]
In the exercises, from 9 to 11, find the domain of the function.
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\[ f(x) = \sqrt{4-x} + \sqrt{x-4} \]
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\[ f(x) = \sqrt{-x} + \cfrac{1}{\sqrt{x+2}} \]
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\[ g(x) = \cfrac{ \sqrt{3-x} + \sqrt{x+2} }{ x^2-9 } \]
In the exercises, from 12 to 16, find \(\boldsymbol{f \circ g,\,g \circ f,\,f \circ f}\) and \(\boldsymbol{g \circ g}\), with their respective domains.
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\[ f(x) = x^2-1,\; g(x) = \sqrt{x} \]
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\[ f(x) = x^2,\; g(x) = \sqrt{x-4} \]
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\[ f(x) = x^2-x,\; g(x) = \cfrac{1}{x} \]
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\[ f(x) = \cfrac{1}{1-x},\; g(x) = \sqrt[3]{x} \]
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\[ f(x) = \sqrt{x^2-1},\; g(x) = \sqrt{1-x} \]
In the exercises, from 17 to 18, find \(\boldsymbol{f \circ g \circ h}\).
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\(f(x) = \sqrt{x}\), \(g(x) = \cfrac{1}{x}\), \(h(x) = x^2-1\)
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\(f(x) = \sqrt[3]{x}\), \(g(x) = \cfrac{x}{1+x}\), \(h(x) = x^2-x\)
- If \(f(x)=\cfrac{1}{1-x}\), find \(f \circ f \circ f\) with its domain.
In the exercises, from 20 to 23, find two functions \(\boldsymbol{f}\) and \(\boldsymbol{g}\) such that \(\boldsymbol{F=f \circ g}\)}.
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\[ F(x) = \cfrac{1}{1+x} \]
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\[ F(x) = -3 + \sqrt{x} \]
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\[ F(x) = \sqrt[3]{ (2x-1)^2 } \]
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\[ F(x) = \cfrac{1}{ \sqrt{ x^2 - x + 1 } } \]
In the exercises, from 24 to 26, find \(\boldsymbol{f,\, g}\) and \(\boldsymbol{h}\) such that:
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\[ F(x) = \frac{x^2}{1+x^2} \]
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\[ F(x) = \sqrt[3]{ x^2 + \mid x \mid + 1 } \]
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\[ F(x) = \sqrt[4]{ \sqrt{x} - 1 } \]
- If \(f(x)=2x+3\) and \(h(x)=2x^2-4x+5\) find a function \(g\) such that \(f \circ g=h\).
- If \(f(x)=x-3\) and \(h(x)=\cfrac{1}{x-2}\) find a function \(g\) such that \(g \circ f= h\).
Answers
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\(y = x^3 – 3\)
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\(y = (x – 1)^3\)
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\(y = -x^3 + 1\)
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\(-y = (x – 1)^3 + 1\)
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\(y = \cfrac{1}{x} – 2\)
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\(y = \cfrac{1}{x-2}\)
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\(y = – \cfrac{1}{x}\)
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\(y = \cfrac{1}{x – 2} + 5\)
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\(y = – \lfloor x \rfloor\)
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\(y = \lfloor 2x \rfloor\)
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\(y = \cfrac{1}{2} \lfloor x \rfloor \)
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periodo \(= \cfrac{2 \pi}{4} = \cfrac{\pi}{2}\)
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\(\text{Dom}(f + g) = \text{Dom}(f – g)\) \(= \text{Dom}(fg) = (-\infty, \, 1) \cup (1, \, 2]\),
\(\text{Dom}\left( \frac{f}{g} \right) = ( – \infty, \, 1 ) \cup (1, \, 2 )\)
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\(\text{Dom}(f + g) = \text{Dom}(f – g)\) \(= \text{Dom}(fg) = [-4, \, -2] \cup [2, \, 4],\)
\(\text{Dom}\left( \frac{f}{g} \right) = [-4, \, -2)\) \(\cup (2, \, 4]\)
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\(\text{Dom}(f + g) = \text{Dom}(f – g)\) \(= \text{Dom}(fg) = (-2, \, 2)\),
\(\text{Dom}\left( \frac{f}{g} \right) = (-2, \, 2) – \{ 0 \}\)
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\(\text{Dom}(f) = 4\)
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\(\text{Dom}(f) = (-2, \, 0]\)
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\(\text{Dom}(g) = [-2, \, 3 )\)
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\((f \circ g)(x) = x – 1\), \(\text{Dom}(f \circ g) = [0, \, +\infty)\)
\((g \circ f)(x) = \sqrt{ x^2 – 1 }\), \(\text{Dom}(g \circ f) = (-\infty, \, -1] \cup [1, \, +\infty)\)
\((f \circ f)(x) = x^4 – 2x^2\), \(\text{Dom}(f \circ f) = \mathbb{R}\)
\((g \circ g)(x) = \sqrt[4]{x}\), \(\text{Dom}(g \circ g) = [0, \, +\infty)\)
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\((f \circ g)(x) = x – 4\), \(\text{Dom}(f \circ g) = [4, \, +\infty)\),
\((g \circ f)(x) = \sqrt{ x^2 – 4 }\), \(\text{Dom}(g \circ f) = (-\infty, \, -2] \cup [2, \, +\infty)\)
\((f \circ f)(x) = x^4\), \(\text{Dom}(f \circ f) = \mathbb{R}\)
\((g \circ g)(x) = \sqrt{ \sqrt{x – 4} – 4 }\), \(\text{Dom}(g \circ g) = [20, \, +\infty)\)
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\((f \circ g)(x) = \cfrac{1}{x^2} – \cfrac{1}{x}\), \(\text{Dom}(f \circ g) = \mathbb{R}- \{ 0 \}\)
\((g \circ f)(x) = \cfrac{1}{x^2 – x}\), \(\text{Dom}(g \circ f) = \mathbb{R} – \{ 0, \, 1 \}\)
\((f \circ f)(x) = x^4 – 2x^3 + x\), \(\text{Dom}(f \circ f) = \mathbb{R}\)
\((g \circ g)(x) = x\), \(\text{Dom}(g \circ g) = \mathbb{R} – \{ 0 \}\)
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\((f \circ g)(x) = \cfrac{ 1 }{1 – \sqrt[3]{x}}\), \(\text{Dom}(f \circ g) = \mathbb{R} – \{ 1 \}\)
\((g \circ f)(x) = \cfrac{1}{\sqrt[3]{ 1 – x }}\), \(\text{Dom}(g \circ f) = \mathbb{R} – \{ 1 \}\)
\((f \circ f)(x) = \cfrac{x – 1}{x}\), \(\text{Dom}(f \circ f) = \mathbb{R} – \{ 0, \, 1\}\)
\((g \circ g)(x) = \sqrt[9]{x}\), \(\text{Dom}(g \circ g) = \mathbb{R}\)
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\((f \circ g)(x) = \sqrt{-x}\), \(\text{Dom}(f \circ g) =(-\infty, \, 0]\)
\((g \circ f)(x) = \sqrt{ 1 – \sqrt{ x^2 – 1 } }\), \(\text{Dom}(g \circ f) = \left[ -\sqrt{2}, \, -1 \right]\) \(\cup \left[ 1, \, \sqrt{2} \right] \)
\((f \circ f)(x) = \sqrt{x^2 – 2}\), \(\text{Dom}(f \circ f) = \left( -\infty, \, -\sqrt{2} \right]\) \(\cup \left[ \sqrt{2}, \, +\infty \right)\)
\((g \circ g)(x) = \sqrt{ 1 – \sqrt{ 1 – x } }\), \(\text{Dom}(g \circ g) = [0, \, 1]\)
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\[ (f \circ g \circ h)(x) = \sqrt{ \frac{1}{x^2 – 1} } \]
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\[ (f \circ g \circ h)(x) = \sqrt[3]{ \frac{x^2 – x}{ x^2 – x + 1 } } \]
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\((f \circ f \circ f)(x) = x, \) \(\text{Dom}(f \circ f \circ f) = \mathbb{R}- \{ 0, \, 1 \}\)
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\[ f(x) = \frac{1}{x}, \, g(x) = 1 + x \]
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\[ f(x) = x-3, \, g(x) = \sqrt{x} \]
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\[ f(x) = \sqrt[3]{x}, \, g(x) = (2x – 1)^2 \]
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\[ f(x) = \frac{1}{x}, \, g(x) = \sqrt{ x^2 – x + 1 } \]
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\(f(x) = \cfrac{1}{x + 1}\), \(g(x) = \cfrac{1}{x}\), \(h(x) = x^2\)
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\(f(x) = \sqrt[3]{x}\), \(g(x) = x + 1\), \(h(x) = x^2 + \mid x \mid\)
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\(f(x) = \sqrt[4]{x}\), \(g(x) = x – 1\), \(h(x) = \sqrt{x}\)
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\[ g(x) = x^2 – 2x + 1 \]
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\[ g(x) = \frac{1}{x + 1} \]