Question

la fonction valeur absolue
1 dans chaque cas, déterminez la règle de la fonction définie par parties dont la représentation graphique correspond à celle de la fonction valeur absolue.
a) ( f(x) = 5|x + 1| - 3 )
b) ( g(x) = -3|x - 4| + 2 )
c) ( h(x) = 0,1|x + 7| - 1 )

Explanation:

To determine the piece - wise function for an absolute - value function of the form \(y = a|x - h|+k\), we use the definition of the absolute value: \(|u|=

$$\begin{cases}u, & u\geq0\\-u, & u < 0\end{cases}$$

\). We first find the critical point by setting the expression inside the absolute value equal to zero (\(x - h=0\) gives \(x = h\)), and then we consider the two cases \(x\geq h\) and \(x < h\).

Part (a): For \(f(x)=5|x + 1|-3\)

Step 1: Find the critical point

Set \(x + 1 = 0\). Solving for \(x\), we get \(x=-1\).

Step 2: Case 1: \(x\geq - 1\)

When \(x\geq - 1\), \(x + 1\geq0\), so \(|x + 1|=x + 1\).
Substitute \(|x + 1|=x + 1\) into \(f(x)\):
\(f(x)=5(x + 1)-3\)
Expand the expression: \(f(x)=5x+5 - 3=5x + 2\)

Step 3: Case 2: \(x < - 1\)

When \(x < - 1\), \(x + 1<0\), so \(|x + 1|=-(x + 1)=-x - 1\).
Substitute \(|x + 1|=-x - 1\) into \(f(x)\):
\(f(x)=5(-x - 1)-3\)
Expand the expression: \(f(x)=-5x-5 - 3=-5x - 8\)

So, \(f(x)=

$$\begin{cases}5x + 2, & x\geq - 1\\-5x - 8, & x < - 1\end{cases}$$

\)

Part (b): For \(g(x)=-3|x - 4|+2\)

Step 1: Find the critical point

Set \(x - 4 = 0\). Solving for \(x\), we get \(x = 4\).

Step 2: Case 1: \(x\geq4\)

When \(x\geq4\), \(x - 4\geq0\), so \(|x - 4|=x - 4\).
Substitute \(|x - 4|=x - 4\) into \(g(x)\):
\(g(x)=-3(x - 4)+2\)
Expand the expression: \(g(x)=-3x + 12+2=-3x+14\)

Step 3: Case 2: \(x < 4\)

When \(x < 4\), \(x - 4<0\), so \(|x - 4|=-(x - 4)=-x + 4\).
Substitute \(|x - 4|=-x + 4\) into \(g(x)\):
\(g(x)=-3(-x + 4)+2\)
Expand the expression: \(g(x)=3x-12 + 2=3x-10\)

So, \(g(x)=

$$\begin{cases}-3x + 14, & x\geq4\\3x - 10, & x < 4\end{cases}$$

\)

Part (c): For \(h(x)=0.1|x + 7|-1\)

Step 1: Find the critical point

Set \(x + 7 = 0\). Solving for \(x\), we get \(x=-7\).

Step 2: Case 1: \(x\geq - 7\)

When \(x\geq - 7\), \(x + 7\geq0\), so \(|x + 7|=x + 7\).
Substitute \(|x + 7|=x + 7\) into \(h(x)\):
\(h(x)=0.1(x + 7)-1\)
Expand the expression: \(h(x)=0.1x+0.7 - 1=0.1x - 0.3\)

Step 3: Case 2: \(x < - 7\)

When \(x < - 7\), \(x + 7<0\), so \(|x + 7|=-(x + 7)=-x - 7\).
Substitute \(|x + 7|=-x - 7\) into \(h(x)\):
\(h(x)=0.1(-x - 7)-1\)
Expand the expression: \(h(x)=-0.1x-0.7 - 1=-0.1x - 1.7\)

So, \(h(x)=

$$\begin{cases}0.1x - 0.3, & x\geq - 7\\-0.1x - 1.7, & x < - 7\end{cases}$$

\)

Final Answers

a) \(\boldsymbol{f(x)=

$$\begin{cases}5x + 2, & x\geq - 1\\-5x - 8, & x < - 1\end{cases}$$

}\)

b) \(\boldsymbol{g(x)=

$$\begin{cases}-3x + 14, & x\geq4\\3x - 10, & x < 4\end{cases}$$

}\)

c) \(\boldsymbol{h(x)=

$$\begin{cases}0.1x - 0.3, & x\geq - 7\\-0.1x - 1.7, & x < - 7\end{cases}$$

}\)

Answer:

To determine the piece - wise function for an absolute - value function of the form \(y = a|x - h|+k\), we use the definition of the absolute value: \(|u|=

$$\begin{cases}u, & u\geq0\\-u, & u < 0\end{cases}$$

\). We first find the critical point by setting the expression inside the absolute value equal to zero (\(x - h=0\) gives \(x = h\)), and then we consider the two cases \(x\geq h\) and \(x < h\).

Part (a): For \(f(x)=5|x + 1|-3\)

Step 1: Find the critical point

Set \(x + 1 = 0\). Solving for \(x\), we get \(x=-1\).

Step 2: Case 1: \(x\geq - 1\)

When \(x\geq - 1\), \(x + 1\geq0\), so \(|x + 1|=x + 1\).
Substitute \(|x + 1|=x + 1\) into \(f(x)\):
\(f(x)=5(x + 1)-3\)
Expand the expression: \(f(x)=5x+5 - 3=5x + 2\)

Step 3: Case 2: \(x < - 1\)

When \(x < - 1\), \(x + 1<0\), so \(|x + 1|=-(x + 1)=-x - 1\).
Substitute \(|x + 1|=-x - 1\) into \(f(x)\):
\(f(x)=5(-x - 1)-3\)
Expand the expression: \(f(x)=-5x-5 - 3=-5x - 8\)

So, \(f(x)=

$$\begin{cases}5x + 2, & x\geq - 1\\-5x - 8, & x < - 1\end{cases}$$

\)

Part (b): For \(g(x)=-3|x - 4|+2\)

Step 1: Find the critical point

Set \(x - 4 = 0\). Solving for \(x\), we get \(x = 4\).

Step 2: Case 1: \(x\geq4\)

When \(x\geq4\), \(x - 4\geq0\), so \(|x - 4|=x - 4\).
Substitute \(|x - 4|=x - 4\) into \(g(x)\):
\(g(x)=-3(x - 4)+2\)
Expand the expression: \(g(x)=-3x + 12+2=-3x+14\)

Step 3: Case 2: \(x < 4\)

When \(x < 4\), \(x - 4<0\), so \(|x - 4|=-(x - 4)=-x + 4\).
Substitute \(|x - 4|=-x + 4\) into \(g(x)\):
\(g(x)=-3(-x + 4)+2\)
Expand the expression: \(g(x)=3x-12 + 2=3x-10\)

So, \(g(x)=

$$\begin{cases}-3x + 14, & x\geq4\\3x - 10, & x < 4\end{cases}$$

\)

Part (c): For \(h(x)=0.1|x + 7|-1\)

Step 1: Find the critical point

Set \(x + 7 = 0\). Solving for \(x\), we get \(x=-7\).

Step 2: Case 1: \(x\geq - 7\)

When \(x\geq - 7\), \(x + 7\geq0\), so \(|x + 7|=x + 7\).
Substitute \(|x + 7|=x + 7\) into \(h(x)\):
\(h(x)=0.1(x + 7)-1\)
Expand the expression: \(h(x)=0.1x+0.7 - 1=0.1x - 0.3\)

Step 3: Case 2: \(x < - 7\)

When \(x < - 7\), \(x + 7<0\), so \(|x + 7|=-(x + 7)=-x - 7\).
Substitute \(|x + 7|=-x - 7\) into \(h(x)\):
\(h(x)=0.1(-x - 7)-1\)
Expand the expression: \(h(x)=-0.1x-0.7 - 1=-0.1x - 1.7\)

So, \(h(x)=

$$\begin{cases}0.1x - 0.3, & x\geq - 7\\-0.1x - 1.7, & x < - 7\end{cases}$$

\)

Final Answers

a) \(\boldsymbol{f(x)=

$$\begin{cases}5x + 2, & x\geq - 1\\-5x - 8, & x < - 1\end{cases}$$

}\)

b) \(\boldsymbol{g(x)=

$$\begin{cases}-3x + 14, & x\geq4\\3x - 10, & x < 4\end{cases}$$

}\)

c) \(\boldsymbol{h(x)=

$$\begin{cases}0.1x - 0.3, & x\geq - 7\\-0.1x - 1.7, & x < - 7\end{cases}$$

}\)