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Sagot :
Certainly! Let's tackle each part of the problem step by step.
(a) State the value of [tex]\( x \)[/tex] that cannot be included in any domain of the function [tex]\( f \)[/tex].
The given function is:
[tex]\[ f(x) = \frac{3x + 1}{x - 2} \][/tex]
To determine which [tex]\( x \)[/tex]-values cannot be included in the domain, we need to identify values that make the denominator zero, as division by zero is undefined.
Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, [tex]\( x = 2 \)[/tex] cannot be included in the domain of the function [tex]\( f \)[/tex]. Thus, the value of [tex]\( x \)[/tex] that cannot be included in the domain is [tex]\( x = 2 \)[/tex].
(b) Express the inverse function [tex]\( f^{-1} \)[/tex] in the form [tex]\( f^{-1}(x) = \ldots \)[/tex]
To find the inverse of the function [tex]\( f \)[/tex], we start with the equation:
[tex]\[ y = \frac{3x + 1}{x - 2} \][/tex]
We want to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. Follow these steps:
1. Swap [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3x + 1}{x - 2} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
First, multiply both sides by [tex]\( x - 2 \)[/tex] to eliminate the denominator:
[tex]\[ y(x - 2) = 3x + 1 \][/tex]
Distribute [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ yx - 2y = 3x + 1 \][/tex]
Collect all [tex]\( x \)[/tex]-terms on one side of the equation and constants on the other:
[tex]\[ yx - 3x = 2y + 1 \][/tex]
Factor out [tex]\( x \)[/tex] on the left-hand side:
[tex]\[ x(y - 3) = 2y + 1 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2y + 1}{y - 3} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 3} \][/tex]
So, in summary:
(a) The value of [tex]\( x \)[/tex] that cannot be included in the domain of the function [tex]\( f \)[/tex] is [tex]\( x = 2 \)[/tex].
(b) The inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 3} \][/tex]
(a) State the value of [tex]\( x \)[/tex] that cannot be included in any domain of the function [tex]\( f \)[/tex].
The given function is:
[tex]\[ f(x) = \frac{3x + 1}{x - 2} \][/tex]
To determine which [tex]\( x \)[/tex]-values cannot be included in the domain, we need to identify values that make the denominator zero, as division by zero is undefined.
Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, [tex]\( x = 2 \)[/tex] cannot be included in the domain of the function [tex]\( f \)[/tex]. Thus, the value of [tex]\( x \)[/tex] that cannot be included in the domain is [tex]\( x = 2 \)[/tex].
(b) Express the inverse function [tex]\( f^{-1} \)[/tex] in the form [tex]\( f^{-1}(x) = \ldots \)[/tex]
To find the inverse of the function [tex]\( f \)[/tex], we start with the equation:
[tex]\[ y = \frac{3x + 1}{x - 2} \][/tex]
We want to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. Follow these steps:
1. Swap [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3x + 1}{x - 2} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
First, multiply both sides by [tex]\( x - 2 \)[/tex] to eliminate the denominator:
[tex]\[ y(x - 2) = 3x + 1 \][/tex]
Distribute [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ yx - 2y = 3x + 1 \][/tex]
Collect all [tex]\( x \)[/tex]-terms on one side of the equation and constants on the other:
[tex]\[ yx - 3x = 2y + 1 \][/tex]
Factor out [tex]\( x \)[/tex] on the left-hand side:
[tex]\[ x(y - 3) = 2y + 1 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2y + 1}{y - 3} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 3} \][/tex]
So, in summary:
(a) The value of [tex]\( x \)[/tex] that cannot be included in the domain of the function [tex]\( f \)[/tex] is [tex]\( x = 2 \)[/tex].
(b) The inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 3} \][/tex]
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