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Sagot :
Sure! Let's address each part of the problem step-by-step.
### Part (a): Find [tex]\( f^{-1}(x) \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x) = 7x - 6 \)[/tex]:
1. Start by writing the function [tex]\( y = 7x - 6 \)[/tex].
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = 7y - 6 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x = 7y - 6 \][/tex]
Add 6 to both sides:
[tex]\[ x + 6 = 7y \][/tex]
Divide both sides by 7:
[tex]\[ y = \frac{x + 6}{7} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 6}{7} \][/tex]
### Part (b): Graph [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] in the same rectangular coordinate system
To graph [tex]\( f(x) = 7x - 6 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex], follow these steps:
1. Choose values for [tex]\( x \)[/tex] and calculate corresponding [tex]\( y \)[/tex]-values for both functions.
2. Plot those points and draw the lines.
For [tex]\( f(x) = 7x - 6 \)[/tex]:
- If [tex]\( x = -2 \)[/tex], then [tex]\( y = 7(-2) - 6 = -14 - 6 = -20 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = 7(0) - 6 = -6 \)[/tex]
- If [tex]\( x = 2 \)[/tex], then [tex]\( y = 7(2) - 6 = 14 - 6 = 8 \)[/tex]
For [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex]:
- If [tex]\( x = -14 \)[/tex], then [tex]\( y = \frac{-14 + 6}{7} = \frac{-8}{7} \approx -1.14 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = \frac{0 + 6}{7} = \frac{6}{7} \approx 0.86 \)[/tex]
- If [tex]\( x = 14 \)[/tex], then [tex]\( y = \frac{14 + 6}{7} = \frac{20}{7} \approx 2.86 \)[/tex]
### Part (c): Use interval notation to give the domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
For both [tex]\( f(x) = 7x - 6 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex]:
- The domain is all real numbers ([tex]\( (-\infty, \infty) \)[/tex]).
- The range is also all real numbers ([tex]\( (-\infty, \infty) \)[/tex]).
Therefore, the answers are:
- Domain of [tex]\( f \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Range of [tex]\( f \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Domain of [tex]\( f^{-1} \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Range of [tex]\( f^{-1} \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
### Part (a): Find [tex]\( f^{-1}(x) \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x) = 7x - 6 \)[/tex]:
1. Start by writing the function [tex]\( y = 7x - 6 \)[/tex].
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = 7y - 6 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x = 7y - 6 \][/tex]
Add 6 to both sides:
[tex]\[ x + 6 = 7y \][/tex]
Divide both sides by 7:
[tex]\[ y = \frac{x + 6}{7} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 6}{7} \][/tex]
### Part (b): Graph [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] in the same rectangular coordinate system
To graph [tex]\( f(x) = 7x - 6 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex], follow these steps:
1. Choose values for [tex]\( x \)[/tex] and calculate corresponding [tex]\( y \)[/tex]-values for both functions.
2. Plot those points and draw the lines.
For [tex]\( f(x) = 7x - 6 \)[/tex]:
- If [tex]\( x = -2 \)[/tex], then [tex]\( y = 7(-2) - 6 = -14 - 6 = -20 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = 7(0) - 6 = -6 \)[/tex]
- If [tex]\( x = 2 \)[/tex], then [tex]\( y = 7(2) - 6 = 14 - 6 = 8 \)[/tex]
For [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex]:
- If [tex]\( x = -14 \)[/tex], then [tex]\( y = \frac{-14 + 6}{7} = \frac{-8}{7} \approx -1.14 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = \frac{0 + 6}{7} = \frac{6}{7} \approx 0.86 \)[/tex]
- If [tex]\( x = 14 \)[/tex], then [tex]\( y = \frac{14 + 6}{7} = \frac{20}{7} \approx 2.86 \)[/tex]
### Part (c): Use interval notation to give the domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
For both [tex]\( f(x) = 7x - 6 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 6}{7} \)[/tex]:
- The domain is all real numbers ([tex]\( (-\infty, \infty) \)[/tex]).
- The range is also all real numbers ([tex]\( (-\infty, \infty) \)[/tex]).
Therefore, the answers are:
- Domain of [tex]\( f \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Range of [tex]\( f \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Domain of [tex]\( f^{-1} \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
- Range of [tex]\( f^{-1} \)[/tex]: [tex]\((- \infty, \infty)\)[/tex]
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