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Sagot :
To determine the intervals where the curve [tex]\( y = -3 x^{\frac{2}{3}} (x-5) \)[/tex] is increasing, we need to:
1. Find the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
2. Identify the critical points by setting the first derivative equal to zero and solving for [tex]\( x \)[/tex].
3. Analyze the sign of the first derivative around these critical points to determine where the function is increasing (i.e., where the first derivative is positive).
Given the function [tex]\( y = -3 x^{\frac{2}{3}} (x-5) \)[/tex], the first derivative [tex]\( \frac{dy}{dx} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = -2.0 \frac{(x - 5)}{x^{1/3}} - 3 x^{2/3} \][/tex]
Now, let's determine the critical points by setting [tex]\( \frac{dy}{dx} \)[/tex] to zero and solving for [tex]\( x \)[/tex]:
[tex]\[ -2.0 \frac{(x - 5)}{x^{1/3}} - 3 x^{2/3} = 0 \][/tex]
Solving this equation, we find one critical point at [tex]\( x = 2 \)[/tex].
Next, we analyze the sign of [tex]\( \frac{dy}{dx} \)[/tex] around this critical point:
1. For [tex]\( x < 0 \)[/tex], let's choose a test point like [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = -1 \approx -2.0 \frac{(-1 - 5)}{(-1)^{1/3}} - 3 (-1)^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-6)}{-1} - 3 \cdot 1 = -12 - 3 = -15 \][/tex]
Thus, [tex]\( \frac{dy}{dx} < 0 \)[/tex].
2. For [tex]\( 0 < x < 2 \)[/tex], let's choose a test point like [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = 1 \approx -2.0 \frac{(1 - 5)}{1^{1/3}} - 3 \cdot 1^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-4)}{1} - 3 \cdot 1 = 8 - 3 = 5 \][/tex]
Thus, [tex]\( \frac{dy}{dx} > 0 \)[/tex].
3. For [tex]\( x > 2 \)[/tex], let's choose a test point like [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = 3 \approx -2.0 \frac{(3 - 5)}{3^{1/3}} - 3 \cdot 3^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-2)}{3^{1/3}} - 3 \cdot 3^{2/3} = \frac{4}{3^{1/3}} - 9 \cdot 3^{2/3} \][/tex]
Evaluating these terms, we find that [tex]\( \frac{dy}{dx} < 0 \)[/tex].
From this analysis, we find that the function is increasing when [tex]\( 0 < x < 2 \)[/tex]. Therefore, the correct answer is:
(B) [tex]\( 0 < x < 2 \)[/tex].
1. Find the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
2. Identify the critical points by setting the first derivative equal to zero and solving for [tex]\( x \)[/tex].
3. Analyze the sign of the first derivative around these critical points to determine where the function is increasing (i.e., where the first derivative is positive).
Given the function [tex]\( y = -3 x^{\frac{2}{3}} (x-5) \)[/tex], the first derivative [tex]\( \frac{dy}{dx} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = -2.0 \frac{(x - 5)}{x^{1/3}} - 3 x^{2/3} \][/tex]
Now, let's determine the critical points by setting [tex]\( \frac{dy}{dx} \)[/tex] to zero and solving for [tex]\( x \)[/tex]:
[tex]\[ -2.0 \frac{(x - 5)}{x^{1/3}} - 3 x^{2/3} = 0 \][/tex]
Solving this equation, we find one critical point at [tex]\( x = 2 \)[/tex].
Next, we analyze the sign of [tex]\( \frac{dy}{dx} \)[/tex] around this critical point:
1. For [tex]\( x < 0 \)[/tex], let's choose a test point like [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = -1 \approx -2.0 \frac{(-1 - 5)}{(-1)^{1/3}} - 3 (-1)^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-6)}{-1} - 3 \cdot 1 = -12 - 3 = -15 \][/tex]
Thus, [tex]\( \frac{dy}{dx} < 0 \)[/tex].
2. For [tex]\( 0 < x < 2 \)[/tex], let's choose a test point like [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = 1 \approx -2.0 \frac{(1 - 5)}{1^{1/3}} - 3 \cdot 1^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-4)}{1} - 3 \cdot 1 = 8 - 3 = 5 \][/tex]
Thus, [tex]\( \frac{dy}{dx} > 0 \)[/tex].
3. For [tex]\( x > 2 \)[/tex], let's choose a test point like [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{dy}{dx} \text{ at } x = 3 \approx -2.0 \frac{(3 - 5)}{3^{1/3}} - 3 \cdot 3^{2/3} \][/tex]
[tex]\[ = -2.0 \frac{(-2)}{3^{1/3}} - 3 \cdot 3^{2/3} = \frac{4}{3^{1/3}} - 9 \cdot 3^{2/3} \][/tex]
Evaluating these terms, we find that [tex]\( \frac{dy}{dx} < 0 \)[/tex].
From this analysis, we find that the function is increasing when [tex]\( 0 < x < 2 \)[/tex]. Therefore, the correct answer is:
(B) [tex]\( 0 < x < 2 \)[/tex].
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