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Sagot :
To find the domain and range of the function [tex]\( g(x) = -\frac{1}{3} \sqrt{(x-5)(x-9)} + 1 \)[/tex], follow these steps:
### Domain
For the function to be real-valued, the expression inside the square root, [tex]\((x-5)(x-9)\)[/tex], must be non-negative. This leads us to solve the inequality:
[tex]\[ (x-5)(x-9) \geq 0 \][/tex]
To solve this, consider the critical points where the expression is zero, namely [tex]\( x = 5 \)[/tex] and [tex]\( x = 9 \)[/tex]. Between these points, the sign of the product [tex]\((x-5)(x-9)\)[/tex] determines whether the inequality holds.
- For [tex]\( x < 5 \)[/tex], both [tex]\( (x-5) \)[/tex] and [tex]\( (x-9) \)[/tex] are negative, thus their product is positive.
- For [tex]\( 5 \leq x \leq 9 \)[/tex], [tex]\((x-5)\)[/tex] is non-negative and [tex]\((x-9)\)[/tex] is non-positive, making their product negative.
- For [tex]\( x > 9 \)[/tex], both [tex]\((x-5)\)[/tex] and [tex]\((x-9)\)[/tex] are positive, thus their product is positive.
Therefore, the inequality holds for:
[tex]\[ x \in (-\infty, 5] \cup [9, \infty) \][/tex]
Thus, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Domain}: (-\infty, 5] \cup [9, \infty) \][/tex]
### Range
To determine the range, consider the value output by the function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = -\frac{1}{3} \sqrt{(x-5)(x-9)} + 1 \][/tex]
First, note that the term [tex]\(\sqrt{(x-5)(x-9)}\)[/tex] is a non-negative quantity for [tex]\( x \)[/tex] in the domain, reaching its maximum at the boundary points of the intervals. This means:
- At [tex]\( x = 5 \)[/tex]: [tex]\(\sqrt{(x-5)(x-9)} = \sqrt{0} = 0\)[/tex]
- At [tex]\( x = 9 \)[/tex]: [tex]\(\sqrt{(x-5)(x-9)} = \sqrt{0} = 0\)[/tex]
For values within the intervals, the function [tex]\(\sqrt{(x-5)(x-9)}\)[/tex] increases, making [tex]\(-\frac{1}{3} \sqrt{(x-5)(x-9)}\)[/tex] more negative.
The maximum value occurs when [tex]\(\sqrt{(x-5)(x-9)} = 0\)[/tex], which leads to:
[tex]\[ g(x) = -\frac{1}{3} \cdot 0 + 1 = 1 \][/tex]
The range extends from the most negative value of [tex]\(-\frac{1}{3} \sqrt{(x-5)(x-9)}\)[/tex], which is approached as [tex]\((x-5)(x-9)\)[/tex] grows larger, approaching negative infinity.
Therefore, the range of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Range}: \left[\frac{2}{3}, 1\right] \][/tex]
In summary:
- Domain: [tex]\( (-\infty, 5] \cup [9, \infty) \)[/tex]
- Range: [tex]\(\left[\frac{2}{3}, 1\right]\)[/tex]
### Domain
For the function to be real-valued, the expression inside the square root, [tex]\((x-5)(x-9)\)[/tex], must be non-negative. This leads us to solve the inequality:
[tex]\[ (x-5)(x-9) \geq 0 \][/tex]
To solve this, consider the critical points where the expression is zero, namely [tex]\( x = 5 \)[/tex] and [tex]\( x = 9 \)[/tex]. Between these points, the sign of the product [tex]\((x-5)(x-9)\)[/tex] determines whether the inequality holds.
- For [tex]\( x < 5 \)[/tex], both [tex]\( (x-5) \)[/tex] and [tex]\( (x-9) \)[/tex] are negative, thus their product is positive.
- For [tex]\( 5 \leq x \leq 9 \)[/tex], [tex]\((x-5)\)[/tex] is non-negative and [tex]\((x-9)\)[/tex] is non-positive, making their product negative.
- For [tex]\( x > 9 \)[/tex], both [tex]\((x-5)\)[/tex] and [tex]\((x-9)\)[/tex] are positive, thus their product is positive.
Therefore, the inequality holds for:
[tex]\[ x \in (-\infty, 5] \cup [9, \infty) \][/tex]
Thus, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Domain}: (-\infty, 5] \cup [9, \infty) \][/tex]
### Range
To determine the range, consider the value output by the function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = -\frac{1}{3} \sqrt{(x-5)(x-9)} + 1 \][/tex]
First, note that the term [tex]\(\sqrt{(x-5)(x-9)}\)[/tex] is a non-negative quantity for [tex]\( x \)[/tex] in the domain, reaching its maximum at the boundary points of the intervals. This means:
- At [tex]\( x = 5 \)[/tex]: [tex]\(\sqrt{(x-5)(x-9)} = \sqrt{0} = 0\)[/tex]
- At [tex]\( x = 9 \)[/tex]: [tex]\(\sqrt{(x-5)(x-9)} = \sqrt{0} = 0\)[/tex]
For values within the intervals, the function [tex]\(\sqrt{(x-5)(x-9)}\)[/tex] increases, making [tex]\(-\frac{1}{3} \sqrt{(x-5)(x-9)}\)[/tex] more negative.
The maximum value occurs when [tex]\(\sqrt{(x-5)(x-9)} = 0\)[/tex], which leads to:
[tex]\[ g(x) = -\frac{1}{3} \cdot 0 + 1 = 1 \][/tex]
The range extends from the most negative value of [tex]\(-\frac{1}{3} \sqrt{(x-5)(x-9)}\)[/tex], which is approached as [tex]\((x-5)(x-9)\)[/tex] grows larger, approaching negative infinity.
Therefore, the range of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Range}: \left[\frac{2}{3}, 1\right] \][/tex]
In summary:
- Domain: [tex]\( (-\infty, 5] \cup [9, \infty) \)[/tex]
- Range: [tex]\(\left[\frac{2}{3}, 1\right]\)[/tex]
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