Definition

Samia CHALLAL

27 Definition

A function $f$ is a rule that assigns to each element $x$ in a set $D_f$ exactly one element $f(x) \in R_f$ .

$D_f$ is called the domain of $f$ : $D_f =\{ x\in \mathbb{R} : \, \, f(x) \in \mathbb{R}\}$

$R_f$ is called the range of $f$ : $R_f= \{ f(x)\in \mathbb{R} : \, \, x \in D_f\}$

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Exercise 1

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We have

$f( {5}) = {2} | {5} - {5}|= 0$

$f( {-2}) = {2} | {-2} - {5}|= 2\times 7 = {14}$

$f( \frac{1}{{2}}) = {2} |\frac{1}{{2}} - {5}| = {2} |\frac{1 - ({2})({5})}{{2}}| = {2} |\frac{-9}{{2}}|= {9}$

$f( x+1) = {2} | x+1 - {5}| = {2} | x - {4}|$

$f( x^2+1) = {2} | x^2+1 - {5}| = {2} | x^2 - {4}|.$

Exercise 2

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We have

$f( {-4}) ={6} ({-4}) = {-24}$

$f( 0) = x+1\Big|_{x=0} = 1$

$f({3}) = x+1\Big|_{x={3}} = {4}$

$f({4} ) = x+1\Big|_{x=4} = {4} + 1= {5}$

$f( {8}) =( x- {4})^2\Big|_{x={8}} = ({8} - {4})^2= {16}.$

Exercise 3

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We have

$\displaystyle{f( {3}) = \frac{ {6} - {7} ({3})}{{6} + {7} ({3})}= {-15}/{27} }$

$\displaystyle{f(\frac{6}{{7}}) = \frac{ {6} - {7} (\frac{6}{{7}})}{{6} + {7} (\frac{6}{{7}})}= 0 }$

$\displaystyle{f(\frac{6}{{14}}) = \frac{ {6} - {7} (\frac{6}{{14}})}{{6} + {7} (\frac{6}{{14}})}= \frac{ {6} - (\frac{6}{2})}{{6} + (\frac{6}{2})} = \frac{(\frac{6}{2}) }{3(\frac{6}{2})} = \frac{1}{3} }$

$\displaystyle{ f((\frac{6}{{7}})x) = \frac{ {6} - {7} (\frac{6}{{7}})(x)}{{6} + {7} (\frac{6}{{7}}(x))}= \frac{ {6} - {6} (x)}{{6} + {6} (x)} = \frac{1-x}{1+x} }$

$\displaystyle{f(\frac{6}{{7}}(a^2+1)) = \frac{ {6} - {7} (\frac{6}{{7}})(a^2+1)}{{6} + {7} (\frac{6}{{7}}(a^2+1))}= \frac{ {6} - {6} (a^2+1)}{{6} + {6} (a^2+1)} = -\frac{a^2}{2+a^2}}.$

Exercise 4

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We have

$\displaystyle{ x^2 + (y - {2})^2= {5} \quad \Longleftrightarrow\quad (y - {2})^2= {5} - x^2 }$

In particular, choosing $\displaystyle{x= \sqrt{\frac{{5} }{2 } }}$ , we obtain $(y - {2})^2= \frac{{5}}{2}$ .

Thus, we assign to $x$ two distinct images $y= {2} \pm \sqrt{\frac{ {5} }{2} }$ .

Therefore, $y$ cannot be a function of $x$ .

Exercise 5

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We have,

$\displaystyle{ {3} x - {9}|y|= 0 \quad \Longleftrightarrow\quad |y|= \frac{{3}}{{9}} x }$

In particular, choosing $\displaystyle{x= \frac{{9}}{{3} } }$ , we obtain $|y|=1$ .

Thus, we assign two distinct images $y= \pm 1$ .

Therefore, $y$ cannot be a function of $x$ .

Exercise 6

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We have

$\displaystyle{ x^2 + {2} y= {8} \quad \Longleftrightarrow\quad y= \frac{{8}-x^2}{{2}} }$

So, for each $x\in \mathbb{R}$ , we assign a unique image $y$ .

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