| pàg. 183 |
| 7 |
A partir de les funcion \(f\) i \(k\), escriu l'expressió algèbrica de les funcions
\(\displaystyle\frac{1}{f}\),
\(\displaystyle\frac{1}{k}\),
\(\displaystyle\frac{1}{-f}\) i
\(-\displaystyle\frac{1}{k}\).
Esbrina'n el domini.
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| 7 |
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\(\left(\displaystyle\frac{1}{f}\right)(x)=\displaystyle\frac{1}{f(x)}=\displaystyle\frac{x+1}{x^2-2}\)
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\(
\begin{array}{rl}
D_{\frac{1}{f}} &=\left\{x\in\mathbb{R}\mid x^2-2\neq0\right\}\\
&=\left\{x\in\mathbb{R}\mid x\neq-\sqrt{2} \land x\neq+\sqrt{2} \right\}\\
&=\mathbb{R}-\left\{-\sqrt{2},+\sqrt{2}\right\}\\
&=(-\infty,-\sqrt{2})\cup(-\sqrt{2},+\sqrt{2})\cup(+\sqrt{2},+\infty)
\end{array}
\)
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\(\left(\displaystyle\frac{1}{k}\right)(x)=\displaystyle\frac{1}{k(x)}=\displaystyle\frac{1}{\sqrt{x-1}}\)
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\(
\begin{array}{rl}
D_{\frac{1}{k}} &=\left\{x\in\mathbb{R}\mid \sqrt{x-1}\neq0\land x-1\geq0\right\}\\
&=\left\{x\in\mathbb{R}\mid x-1>0\right\}\\
&=\left\{x\in\mathbb{R}\mid x>1 \right\}\\
&=(1,+\infty)
\end{array}
\)
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\(\left(\displaystyle\frac{1}{-f}\right)(x)=\displaystyle\frac{1}{-f(x)}=\displaystyle\frac{x+1}{2-x^2}\)
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\(
\begin{array}{rl}
D_{\frac{1}{-f}} &=\left\{x\in\mathbb{R}\mid 2-x^2\neq0\right\}\\
&=\left\{x\in\mathbb{R}\mid x\neq-\sqrt{2} \land x\neq+\sqrt{2} \right\}\\
&=\mathbb{R}-\left\{-\sqrt{2},+\sqrt{2}\right\}\\
&=(-\infty,-\sqrt{2})\cup(-\sqrt{2},+\sqrt{2})\cup(+\sqrt{2},+\infty)\\
&=D_{\frac{1}{f}}
\end{array}
\)
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\(\left(-\displaystyle\frac{1}{k}\right)(x)=-\displaystyle\frac{1}{k(x)}=-\displaystyle\frac{1}{\sqrt{x-1}}\)
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\(
\begin{array}{rl}
D_{-\frac{1}{k}} &=\left\{x\in\mathbb{R}\mid \sqrt{x-1}\neq0\land x-1\geq0\right\}\\
&=\left\{x\in\mathbb{R}\mid x-1>0\right\}\\
&=\left\{x\in\mathbb{R}\mid x>1 \right\}\\
&=(1,+\infty)\\
&=D_{\frac{1}{k}}
\end{array}
\)
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| 8 |
Donades les funcions
\(f(x)=\displaystyle\frac{2x-4}{x+3}\) i
\(g(x)=\displaystyle\frac{x+2}{3x-9}\),
determina l'expressió de les funcions
\(\displaystyle\frac{f}{g}\),
\(\displaystyle\frac{g}{f}\),
\(\displaystyle\frac{\displaystyle\frac{1}{f}}{g}\) i
\(\displaystyle\frac{\displaystyle\frac{1}{g}}{f}\) i esbrina'n el domini.
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| 8 |
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\(
\left(\displaystyle\frac{f}{g}\right)(x)=\displaystyle\frac{f(x)}{g(x)}
=f(x):g(x)
=\displaystyle\frac{2x-4}{x+3}:\displaystyle\frac{x+2}{3x-9}
=\displaystyle\frac{6(x-2)(x-3)}{(x+3)(x+2)}
\)
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\[
\begin{array}{rl}
D_{\frac{f}{g}}&=\left\{x\in\mathbb{R}\mid x\neq-3 \land x\neq-2 \right\}\\
&=\mathbb{R} - \left\{-3,-2\right\}\\
&=(-\infty,-3)\cup(-3,-2)\cup(-2,+\infty)
\end{array}
\]
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\(
\left(\displaystyle\frac{g}{f}\right)(x)=\displaystyle\frac{g(x)}{f(x)}
=g(x):f(x)
=\displaystyle\frac{x+2}{3x-9}:\displaystyle\frac{2x-4}{x+3}
=\displaystyle\frac{x+3)(x+2)}{6(x-2)(x-3)}\)
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\[
\begin{array}{rl}
D_{\frac{g}{f}}&=\left\{x\in\mathbb{R}\mid x\neq2 \land x\neq3 \right\}\\
&=\mathbb{R} - \left\{2,3\right\}\\
&=(-\infty,2)\cup(2,3)\cup(3,+\infty)
\end{array}
\]
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\(
\left(\displaystyle\frac{\displaystyle\frac{1}{f}}{g}\right)(x) =\displaystyle\frac{\displaystyle\frac{1}{f(x)}}{g(x)}
=(1:f(x)):g(x)
=\displaystyle\frac{1}{f(x)\cdot g(x)}
=\displaystyle\frac{(x+3)(3x-9)}{(2x-4)(x+2)}
\)
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\[
\begin{array}{rl}
D_{\frac{\frac{1}{f}}{g}}&=\left\{x\in\mathbb{R}\mid x\neq-2 \land x\neq2 \right\}\\
&=\mathbb{R} - \left\{-2,+2\right\}\\
&=(-\infty,-2)\cup(-2,+2)\cup(+2,+\infty)
\end{array}
\]
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\(
\left(\displaystyle\frac{\displaystyle\frac{1}{g}}{f}\right)(x) =\displaystyle\frac{\displaystyle\frac{1}{g(x)}}{f(x)}
=(1:g(x)):f(x)
=\displaystyle\frac{1}{g(x)\cdot f(x)}
=\displaystyle\frac{(x+3)(3x-9)}{(2x-4)(x+2)}
\)
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\[
\begin{array}{rl}
D_{\frac{\frac{1}{g}}{f}}&=\left\{x\in\mathbb{R}\mid x\neq-2 \land x\neq2 \right\}\\
&=\mathbb{R} - \left\{-2,+2\right\}\\
&=(-\infty,-2)\cup(-2,+2)\cup(+2,+\infty)
\end{array}
\]
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Òbviament:
\[\frac{\displaystyle\frac{1}{f}}{g}=\frac{\displaystyle\frac{1}{g}}{f}=\displaystyle\frac{1}{f\cdot g}\]
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