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Sagot :
Let's systematically examine the table and determine the unknowns step-by-step.
Given:
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & 3 & 4 & 8 \\ \hline a(x) & 0 & 3 & t \\ \hline a(b(x)) & u & 10 & 0 \\ \hline b^{-1}(x) & v & 3 & 4 \\ \hline a(x)-b(x) & w & -5 & 7 \\ \hline \end{array} \][/tex]
### Step 1: Determine [tex]\( t \)[/tex]
When [tex]\( x = 8 \)[/tex], [tex]\( a(b(x)) = 0 \)[/tex] and [tex]\( x = 8 \)[/tex]
From the third column:
- [tex]\( a(x) = t \)[/tex]
- [tex]\( a(b(8)) = 0 \)[/tex]
Given [tex]\( a(b(8)) = 0 \)[/tex] and we know [tex]\( a(x) \)[/tex], [tex]\( t \)[/tex] must be the value that corresponds to [tex]\( x = 8 \)[/tex]. Thus, since [tex]\( a(8) = t \)[/tex], and [tex]\( a(b(x)) = 0 \)[/tex], [tex]\( a(8) = t = 0 \)[/tex].
So,
[tex]\[ t = 0 \][/tex]
### Step 2: Determine [tex]\( u \)[/tex]
For the first column:
- [tex]\( a(x) = 0 \)[/tex]
- [tex]\( a(b(3)) = u \)[/tex]
Given [tex]\( a(3) = 0 \)[/tex], the value of [tex]\( b(3) \)[/tex] corresponds to a value [tex]\( x \)[/tex] for which [tex]\( a(x) = u \)[/tex]. Since [tex]\( a \)[/tex] of something is required to be [tex]\( u \)[/tex] for [tex]\( x = 3 \)[/tex], since [tex]\( b(x) \)[/tex] for [tex]\( x = 3 \)[/tex].
We also know [tex]\( b(x) = 3 \)[/tex], so:
[tex]\[ u = t \][/tex]
So,
[tex]\[ u = 0 \][/tex]
### Step 3: Determine [tex]\( v \)[/tex]
We observe the first column:
- [tex]\( b^{-1}(3) = v \)[/tex]
Given [tex]\( b^{-1}(4) = 3 \)[/tex]:
[tex]\[ b(x) = 3 \equiv t = 0 \][/tex]
So,
[tex]\[ v = 0 \][/tex]
### Step 4: Determine [tex]\( w \)[/tex]
We use the information in the first column:
- [tex]\( a(x) - b(x) = w \)[/tex]
Given,
[tex]\[ w = a(x) - b(x) = 3 - 8 = 0 - 0 = 3 \][/tex]
So,
[tex]\[ w = -15\][/tex]
Final results are:
1. [tex]\( t = 0 \)[/tex]
2. [tex]\( u = 0 \)[/tex]
3. [tex]\( v = v \)[/tex]
4. [tex]\( w = w \)[/tex]
### Complete Table
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & 3 & 4 & 8 \\ \hline a(x) & 0 & 3 & 0 \\ \hline a(b(x)) & 3 & 10 & 0 \\ \hline b^{-1}(x) & 0 & 3 & 4 \\ \hline a(x)-b(x) & 3 & 7 & -3 \\ \hline \end{array} \][/tex]
Given:
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & 3 & 4 & 8 \\ \hline a(x) & 0 & 3 & t \\ \hline a(b(x)) & u & 10 & 0 \\ \hline b^{-1}(x) & v & 3 & 4 \\ \hline a(x)-b(x) & w & -5 & 7 \\ \hline \end{array} \][/tex]
### Step 1: Determine [tex]\( t \)[/tex]
When [tex]\( x = 8 \)[/tex], [tex]\( a(b(x)) = 0 \)[/tex] and [tex]\( x = 8 \)[/tex]
From the third column:
- [tex]\( a(x) = t \)[/tex]
- [tex]\( a(b(8)) = 0 \)[/tex]
Given [tex]\( a(b(8)) = 0 \)[/tex] and we know [tex]\( a(x) \)[/tex], [tex]\( t \)[/tex] must be the value that corresponds to [tex]\( x = 8 \)[/tex]. Thus, since [tex]\( a(8) = t \)[/tex], and [tex]\( a(b(x)) = 0 \)[/tex], [tex]\( a(8) = t = 0 \)[/tex].
So,
[tex]\[ t = 0 \][/tex]
### Step 2: Determine [tex]\( u \)[/tex]
For the first column:
- [tex]\( a(x) = 0 \)[/tex]
- [tex]\( a(b(3)) = u \)[/tex]
Given [tex]\( a(3) = 0 \)[/tex], the value of [tex]\( b(3) \)[/tex] corresponds to a value [tex]\( x \)[/tex] for which [tex]\( a(x) = u \)[/tex]. Since [tex]\( a \)[/tex] of something is required to be [tex]\( u \)[/tex] for [tex]\( x = 3 \)[/tex], since [tex]\( b(x) \)[/tex] for [tex]\( x = 3 \)[/tex].
We also know [tex]\( b(x) = 3 \)[/tex], so:
[tex]\[ u = t \][/tex]
So,
[tex]\[ u = 0 \][/tex]
### Step 3: Determine [tex]\( v \)[/tex]
We observe the first column:
- [tex]\( b^{-1}(3) = v \)[/tex]
Given [tex]\( b^{-1}(4) = 3 \)[/tex]:
[tex]\[ b(x) = 3 \equiv t = 0 \][/tex]
So,
[tex]\[ v = 0 \][/tex]
### Step 4: Determine [tex]\( w \)[/tex]
We use the information in the first column:
- [tex]\( a(x) - b(x) = w \)[/tex]
Given,
[tex]\[ w = a(x) - b(x) = 3 - 8 = 0 - 0 = 3 \][/tex]
So,
[tex]\[ w = -15\][/tex]
Final results are:
1. [tex]\( t = 0 \)[/tex]
2. [tex]\( u = 0 \)[/tex]
3. [tex]\( v = v \)[/tex]
4. [tex]\( w = w \)[/tex]
### Complete Table
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & 3 & 4 & 8 \\ \hline a(x) & 0 & 3 & 0 \\ \hline a(b(x)) & 3 & 10 & 0 \\ \hline b^{-1}(x) & 0 & 3 & 4 \\ \hline a(x)-b(x) & 3 & 7 & -3 \\ \hline \end{array} \][/tex]
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