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Sagot :
Let's solve this problem step by step.
### Part 1: Finding the Derivative
Given the function:
[tex]\[ y = 2x^2 - 8x\sqrt{x} + 8x + 1 \][/tex]
To find the derivative [tex]\(\frac{dy}{dx}\)[/tex]:
1. Rewrite [tex]\( \sqrt{x} \)[/tex] as [tex]\( x^{1/2} \)[/tex].
2. Differentiate each term with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(8x^{3/2}) + \frac{d}{dx}(8x) + \frac{d}{dx}(1) \][/tex]
Performing the differentiation:
[tex]\[ \frac{d}{dx}(2x^2) = 4x \][/tex]
[tex]\[ \frac{d}{dx}(8x^{3/2}) = 8 \cdot \frac{3}{2} x^{1/2} = 12x^{1/2} \][/tex]
[tex]\[ \frac{d}{dx}(8x) = 8 \][/tex]
[tex]\[ \frac{d}{dx}(1) = 0 \][/tex]
So, the derivative is:
[tex]\[ \frac{dy}{dx} = 4x - 12\sqrt{x} + 8 \][/tex]
### Part 2: Finding the Stationary Points
Stationary points occur where [tex]\(\frac{dy}{dx} = 0\)[/tex].
Setting the derivative equal to zero:
[tex]\[ 4x - 12\sqrt{x} + 8 = 0 \][/tex]
To solve this equation, let [tex]\( u = \sqrt{x} \)[/tex]. Then [tex]\( x = u^2 \)[/tex].
Substituting [tex]\( u \)[/tex]:
[tex]\[ 4u^2 - 12u + 8 = 0 \][/tex]
This is a quadratic equation in [tex]\( u \)[/tex]:
[tex]\[ 4u^2 - 12u + 8 = 0 \][/tex]
Dividing by 4 to simplify:
[tex]\[ u^2 - 3u + 2 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (u - 1)(u - 2) = 0 \][/tex]
Thus, [tex]\( u = 1 \)[/tex] or [tex]\( u = 2 \)[/tex]. Converting back to [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x} = 1 \implies x = 1 \][/tex]
[tex]\[ \sqrt{x} = 2 \implies x = 4 \][/tex]
So, the stationary points are at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
To find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^2 - 8(1)\sqrt{1} + 8(1) + 1 = 2 - 8 + 8 + 1 = 3 \][/tex]
For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 2(4)^2 - 8(4)\sqrt{4} + 8(4) + 1 = 2(16) - 8(4)(2) + 32 + 1 = 32 - 64 + 32 + 1 = 1 \][/tex]
Thus, the stationary points are:
[tex]\[ (1, 3) \quad \text{and} \quad (4, 1) \][/tex]
### Part 3: Determining the Nature of Each Stationary Point
To determine the nature of each stationary point, we use the second derivative test. Let's find the second derivative.
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x - 12\sqrt{x} + 8) \][/tex]
Differentiating term by term:
[tex]\[ \frac{d}{dx}(4x) = 4 \][/tex]
[tex]\[ \frac{d}{dx}(-12\sqrt{x}) = -12 \cdot \frac{1}{2} x^{-1/2} = -6x^{-1/2} \][/tex]
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
So, the second derivative is:
[tex]\[ \frac{d^2y}{dx^2} = 4 - 6x^{-1/2} \][/tex]
We now evaluate the second derivative at the stationary points.
For [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{d^2y}{dx^2} \Bigg|_{x=1} = 4 - 6(1)^{-1/2} = 4 - 6 = -2 \][/tex]
Since [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\( x = 1 \)[/tex] is negative, this point is a local maximum.
For [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{d^2y}{dx^2} \Bigg|_{x=4} = 4 - 6(4)^{-1/2} = 4 - 3 = 1 \][/tex]
Since [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\( x = 4 \)[/tex] is positive, this point is a local minimum.
### Summary
1. The derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \frac{dy}{dx} = 4x - 12\sqrt{x} + 8 \][/tex]
2. The stationary points are:
[tex]\[ (1, 3) \quad \text{and} \quad (4, 1) \][/tex]
3. The nature of these points are:
[tex]\[ (1, 3) \quad \text{is a local maximum} \][/tex]
[tex]\[ (4, 1) \quad \text{is a local minimum} \][/tex]
### Part 1: Finding the Derivative
Given the function:
[tex]\[ y = 2x^2 - 8x\sqrt{x} + 8x + 1 \][/tex]
To find the derivative [tex]\(\frac{dy}{dx}\)[/tex]:
1. Rewrite [tex]\( \sqrt{x} \)[/tex] as [tex]\( x^{1/2} \)[/tex].
2. Differentiate each term with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(8x^{3/2}) + \frac{d}{dx}(8x) + \frac{d}{dx}(1) \][/tex]
Performing the differentiation:
[tex]\[ \frac{d}{dx}(2x^2) = 4x \][/tex]
[tex]\[ \frac{d}{dx}(8x^{3/2}) = 8 \cdot \frac{3}{2} x^{1/2} = 12x^{1/2} \][/tex]
[tex]\[ \frac{d}{dx}(8x) = 8 \][/tex]
[tex]\[ \frac{d}{dx}(1) = 0 \][/tex]
So, the derivative is:
[tex]\[ \frac{dy}{dx} = 4x - 12\sqrt{x} + 8 \][/tex]
### Part 2: Finding the Stationary Points
Stationary points occur where [tex]\(\frac{dy}{dx} = 0\)[/tex].
Setting the derivative equal to zero:
[tex]\[ 4x - 12\sqrt{x} + 8 = 0 \][/tex]
To solve this equation, let [tex]\( u = \sqrt{x} \)[/tex]. Then [tex]\( x = u^2 \)[/tex].
Substituting [tex]\( u \)[/tex]:
[tex]\[ 4u^2 - 12u + 8 = 0 \][/tex]
This is a quadratic equation in [tex]\( u \)[/tex]:
[tex]\[ 4u^2 - 12u + 8 = 0 \][/tex]
Dividing by 4 to simplify:
[tex]\[ u^2 - 3u + 2 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (u - 1)(u - 2) = 0 \][/tex]
Thus, [tex]\( u = 1 \)[/tex] or [tex]\( u = 2 \)[/tex]. Converting back to [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x} = 1 \implies x = 1 \][/tex]
[tex]\[ \sqrt{x} = 2 \implies x = 4 \][/tex]
So, the stationary points are at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex].
To find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^2 - 8(1)\sqrt{1} + 8(1) + 1 = 2 - 8 + 8 + 1 = 3 \][/tex]
For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 2(4)^2 - 8(4)\sqrt{4} + 8(4) + 1 = 2(16) - 8(4)(2) + 32 + 1 = 32 - 64 + 32 + 1 = 1 \][/tex]
Thus, the stationary points are:
[tex]\[ (1, 3) \quad \text{and} \quad (4, 1) \][/tex]
### Part 3: Determining the Nature of Each Stationary Point
To determine the nature of each stationary point, we use the second derivative test. Let's find the second derivative.
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x - 12\sqrt{x} + 8) \][/tex]
Differentiating term by term:
[tex]\[ \frac{d}{dx}(4x) = 4 \][/tex]
[tex]\[ \frac{d}{dx}(-12\sqrt{x}) = -12 \cdot \frac{1}{2} x^{-1/2} = -6x^{-1/2} \][/tex]
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
So, the second derivative is:
[tex]\[ \frac{d^2y}{dx^2} = 4 - 6x^{-1/2} \][/tex]
We now evaluate the second derivative at the stationary points.
For [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{d^2y}{dx^2} \Bigg|_{x=1} = 4 - 6(1)^{-1/2} = 4 - 6 = -2 \][/tex]
Since [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\( x = 1 \)[/tex] is negative, this point is a local maximum.
For [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{d^2y}{dx^2} \Bigg|_{x=4} = 4 - 6(4)^{-1/2} = 4 - 3 = 1 \][/tex]
Since [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\( x = 4 \)[/tex] is positive, this point is a local minimum.
### Summary
1. The derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \frac{dy}{dx} = 4x - 12\sqrt{x} + 8 \][/tex]
2. The stationary points are:
[tex]\[ (1, 3) \quad \text{and} \quad (4, 1) \][/tex]
3. The nature of these points are:
[tex]\[ (1, 3) \quad \text{is a local maximum} \][/tex]
[tex]\[ (4, 1) \quad \text{is a local minimum} \][/tex]
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