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Sagot :
To find the inverse [tex]\( h^{-1}(x) \)[/tex] of the given function [tex]\( h(x) = \frac{5x}{x - 5} \)[/tex], and to state its domain and range, we will follow a series of steps. Here’s a detailed solution:
1. Express the function:
[tex]\[ h(x) = \frac{5x}{x - 5} \][/tex]
2. Swap the variables: Since we are looking for the inverse, we start by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] (where [tex]\( y = h(x) \)[/tex]).
[tex]\[ x = \frac{5y}{y - 5} \][/tex]
3. Solve for [tex]\( y \)[/tex]: Rearrange this equation to solve for [tex]\( y \)[/tex].
[tex]\[ x(y - 5) = 5y \][/tex]
Distribute [tex]\( x \)[/tex]:
[tex]\[ xy - 5x = 5y \][/tex]
Bring all terms involving [tex]\( y \)[/tex] to one side of the equation:
[tex]\[ xy - 5y = 5x \][/tex]
Factor out [tex]\( y \)[/tex]:
[tex]\[ y(x - 5) = 5x \][/tex]
Solve for [tex]\( y \)[/tex] by dividing both sides by [tex]\( (x - 5) \)[/tex]:
[tex]\[ y = \frac{5x}{x - 5} \][/tex]
Therefore, the inverse function [tex]\( h^{-1}(x) \)[/tex] is found by replacing [tex]\( y \)[/tex] with [tex]\( x \)[/tex]:
[tex]\[ h^{-1}(x) = \frac{5x}{x - 5} \][/tex]
4. Domain of [tex]\( h(x) \)[/tex]: The function [tex]\( h(x) = \frac{5x}{x - 5} \)[/tex] is undefined when the denominator is zero, i.e., [tex]\( x = 5 \)[/tex]. Thus, the domain of [tex]\( h(x) \)[/tex] is all real numbers except 5:
[tex]\[ \text{Domain of } h(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
5. Range of [tex]\( h(x) \)[/tex]: To find the range of [tex]\( h(x) \)[/tex], notice that it is the set of values that the function can take. By solving the equation [tex]\( y = \frac{5x}{x - 5} \)[/tex], we observe that as [tex]\( x \)[/tex] approaches 5, [tex]\( h(x) \)[/tex] can take any real value except 5. Thus, the range of [tex]\( h(x) \)[/tex] is all real numbers except 5:
[tex]\[ \text{Range of } h(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
6. Domain and range of [tex]\( h^{-1}(x) \)[/tex]: For the inverse function [tex]\( h^{-1}(x) \)[/tex], the domain becomes the range of the original function [tex]\( h(x) \)[/tex], and the range becomes the domain of the original function [tex]\( h(x) \)[/tex].
[tex]\[ \text{Domain of } h^{-1}(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
[tex]\[ \text{Range of } h^{-1}(x): \, (-\infty, 0) \cup (0, \infty) \][/tex]
Therefore, the inverse function and its properties are:
[tex]\[ h^{-1}(x) = \frac{5x}{x - 5} \][/tex]
[tex]\[ \text{Domain of } h^{-1}(x): \, (-\infty, 0) \cup (0, \infty) \][/tex]
[tex]\[ \text{Range of } h^{-1}(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
1. Express the function:
[tex]\[ h(x) = \frac{5x}{x - 5} \][/tex]
2. Swap the variables: Since we are looking for the inverse, we start by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] (where [tex]\( y = h(x) \)[/tex]).
[tex]\[ x = \frac{5y}{y - 5} \][/tex]
3. Solve for [tex]\( y \)[/tex]: Rearrange this equation to solve for [tex]\( y \)[/tex].
[tex]\[ x(y - 5) = 5y \][/tex]
Distribute [tex]\( x \)[/tex]:
[tex]\[ xy - 5x = 5y \][/tex]
Bring all terms involving [tex]\( y \)[/tex] to one side of the equation:
[tex]\[ xy - 5y = 5x \][/tex]
Factor out [tex]\( y \)[/tex]:
[tex]\[ y(x - 5) = 5x \][/tex]
Solve for [tex]\( y \)[/tex] by dividing both sides by [tex]\( (x - 5) \)[/tex]:
[tex]\[ y = \frac{5x}{x - 5} \][/tex]
Therefore, the inverse function [tex]\( h^{-1}(x) \)[/tex] is found by replacing [tex]\( y \)[/tex] with [tex]\( x \)[/tex]:
[tex]\[ h^{-1}(x) = \frac{5x}{x - 5} \][/tex]
4. Domain of [tex]\( h(x) \)[/tex]: The function [tex]\( h(x) = \frac{5x}{x - 5} \)[/tex] is undefined when the denominator is zero, i.e., [tex]\( x = 5 \)[/tex]. Thus, the domain of [tex]\( h(x) \)[/tex] is all real numbers except 5:
[tex]\[ \text{Domain of } h(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
5. Range of [tex]\( h(x) \)[/tex]: To find the range of [tex]\( h(x) \)[/tex], notice that it is the set of values that the function can take. By solving the equation [tex]\( y = \frac{5x}{x - 5} \)[/tex], we observe that as [tex]\( x \)[/tex] approaches 5, [tex]\( h(x) \)[/tex] can take any real value except 5. Thus, the range of [tex]\( h(x) \)[/tex] is all real numbers except 5:
[tex]\[ \text{Range of } h(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
6. Domain and range of [tex]\( h^{-1}(x) \)[/tex]: For the inverse function [tex]\( h^{-1}(x) \)[/tex], the domain becomes the range of the original function [tex]\( h(x) \)[/tex], and the range becomes the domain of the original function [tex]\( h(x) \)[/tex].
[tex]\[ \text{Domain of } h^{-1}(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
[tex]\[ \text{Range of } h^{-1}(x): \, (-\infty, 0) \cup (0, \infty) \][/tex]
Therefore, the inverse function and its properties are:
[tex]\[ h^{-1}(x) = \frac{5x}{x - 5} \][/tex]
[tex]\[ \text{Domain of } h^{-1}(x): \, (-\infty, 0) \cup (0, \infty) \][/tex]
[tex]\[ \text{Range of } h^{-1}(x): \, (-\infty, 5) \cup (5, \infty) \][/tex]
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