At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
