At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Let's address each part of your question in a detailed, step-by-step manner.
### 4) Finding the critical points for [tex]\( f(x) = x^3 - 12x + 5 \)[/tex]
To find the critical points of the function, we need to identify where the first derivative equals zero or does not exist.
1. Compute the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 12x + 5) = 3x^2 - 12. \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 - 12 = 0. \][/tex]
3. Solve the equation:
[tex]\[ 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2. \][/tex]
Thus, the critical points are at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 5) Change of direction for the position function [tex]\( s(t) = t^2 - 7t + 6 \)[/tex] within [tex]\( 0 \leq t \leq 7 \)[/tex]
To determine when the body changes direction, we need to find when the velocity [tex]\( \frac{ds}{dt} \)[/tex] changes sign. This happens when [tex]\( \frac{ds}{dt} = 0 \)[/tex].
1. Compute the first derivative of [tex]\( s(t) \)[/tex]:
[tex]\[ \frac{ds}{dt} = \frac{d}{dt}(t^2 - 7t + 6) = 2t - 7. \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 2t - 7 = 0 \implies t = \frac{7}{2} = 3.5. \][/tex]
3. Check if [tex]\( t = 3.5 \)[/tex] is within the interval [tex]\( 0 \leq t \leq 7 \)[/tex]:
Indeed, [tex]\( 3.5 \)[/tex] is in the interval.
Hence, the body changes direction at [tex]\( t = 3.5 \)[/tex] seconds.
### 6) Derivative of [tex]\( y = \left(\frac{1}{x^2} + 2\right)\left(x^2 - \frac{1}{x^2} + 2\right) \)[/tex]
To find the derivative, we use the product rule [tex]\( (uv)' = u'v + uv' \)[/tex].
1. Let [tex]\( u = \frac{1}{x^2} + 2 \)[/tex] and [tex]\( v = x^2 - \frac{1}{x^2} + 2 \)[/tex].
2. First, compute the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u = \frac{1}{x^2} + 2 \implies u' = -\frac{2}{x^3}, \][/tex]
[tex]\[ v = x^2 - \frac{1}{x^2} + 2 \implies v' = 2x + \frac{2}{x^3}. \][/tex]
3. Apply the product rule:
[tex]\[ y' = u'v + uv' = \left(-\frac{2}{x^3}\right)\left(x^2 - \frac{1}{x^2} + 2\right) + \left(\frac{1}{x^2} + 2\right)\left(2x + \frac{2}{x^3}\right). \][/tex]
Simplify each term separately:
[tex]\[ u'v = \left(-\frac{2}{x^3}\right)\left(x^2 - \frac{1}{x^2} + 2\right) = -\frac{2x^2}{x^3} + \frac{2}{x^5} - \frac{4}{x^3} = -\frac{2}{x} + \frac{2}{x^5} - \frac{4}{x^3}, \][/tex]
[tex]\[ uv' = \left(\frac{1}{x^2} + 2\right)\left(2x + \frac{2}{x^3}\right) = \frac{2x}{x^2} + 4x + \frac{2}{x^5} + \frac{4}{x^3} = \frac{2}{x} + 4x + \frac{2}{x^5} + \frac{4}{x^3}. \][/tex]
Combine both results:
[tex]\[ y' = \left(-\frac{2}{x} + \frac{2}{x^5} - \frac{4}{x^3}\right) + \left(\frac{2}{x} + 4x + \frac{2}{x^5} + \frac{4}{x^3}\right) = 4x + \frac{4}{x^5}. \][/tex]
Thus, the correct derivative is:
[tex]\[ y' = 4x + \frac{4}{x^5}. \][/tex]
So, the correct answer is:
[tex]\[ \boxed{C) \frac{4}{x^5} + 4x} \][/tex]
### 4) Finding the critical points for [tex]\( f(x) = x^3 - 12x + 5 \)[/tex]
To find the critical points of the function, we need to identify where the first derivative equals zero or does not exist.
1. Compute the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 12x + 5) = 3x^2 - 12. \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 - 12 = 0. \][/tex]
3. Solve the equation:
[tex]\[ 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2. \][/tex]
Thus, the critical points are at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 5) Change of direction for the position function [tex]\( s(t) = t^2 - 7t + 6 \)[/tex] within [tex]\( 0 \leq t \leq 7 \)[/tex]
To determine when the body changes direction, we need to find when the velocity [tex]\( \frac{ds}{dt} \)[/tex] changes sign. This happens when [tex]\( \frac{ds}{dt} = 0 \)[/tex].
1. Compute the first derivative of [tex]\( s(t) \)[/tex]:
[tex]\[ \frac{ds}{dt} = \frac{d}{dt}(t^2 - 7t + 6) = 2t - 7. \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 2t - 7 = 0 \implies t = \frac{7}{2} = 3.5. \][/tex]
3. Check if [tex]\( t = 3.5 \)[/tex] is within the interval [tex]\( 0 \leq t \leq 7 \)[/tex]:
Indeed, [tex]\( 3.5 \)[/tex] is in the interval.
Hence, the body changes direction at [tex]\( t = 3.5 \)[/tex] seconds.
### 6) Derivative of [tex]\( y = \left(\frac{1}{x^2} + 2\right)\left(x^2 - \frac{1}{x^2} + 2\right) \)[/tex]
To find the derivative, we use the product rule [tex]\( (uv)' = u'v + uv' \)[/tex].
1. Let [tex]\( u = \frac{1}{x^2} + 2 \)[/tex] and [tex]\( v = x^2 - \frac{1}{x^2} + 2 \)[/tex].
2. First, compute the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u = \frac{1}{x^2} + 2 \implies u' = -\frac{2}{x^3}, \][/tex]
[tex]\[ v = x^2 - \frac{1}{x^2} + 2 \implies v' = 2x + \frac{2}{x^3}. \][/tex]
3. Apply the product rule:
[tex]\[ y' = u'v + uv' = \left(-\frac{2}{x^3}\right)\left(x^2 - \frac{1}{x^2} + 2\right) + \left(\frac{1}{x^2} + 2\right)\left(2x + \frac{2}{x^3}\right). \][/tex]
Simplify each term separately:
[tex]\[ u'v = \left(-\frac{2}{x^3}\right)\left(x^2 - \frac{1}{x^2} + 2\right) = -\frac{2x^2}{x^3} + \frac{2}{x^5} - \frac{4}{x^3} = -\frac{2}{x} + \frac{2}{x^5} - \frac{4}{x^3}, \][/tex]
[tex]\[ uv' = \left(\frac{1}{x^2} + 2\right)\left(2x + \frac{2}{x^3}\right) = \frac{2x}{x^2} + 4x + \frac{2}{x^5} + \frac{4}{x^3} = \frac{2}{x} + 4x + \frac{2}{x^5} + \frac{4}{x^3}. \][/tex]
Combine both results:
[tex]\[ y' = \left(-\frac{2}{x} + \frac{2}{x^5} - \frac{4}{x^3}\right) + \left(\frac{2}{x} + 4x + \frac{2}{x^5} + \frac{4}{x^3}\right) = 4x + \frac{4}{x^5}. \][/tex]
Thus, the correct derivative is:
[tex]\[ y' = 4x + \frac{4}{x^5}. \][/tex]
So, the correct answer is:
[tex]\[ \boxed{C) \frac{4}{x^5} + 4x} \][/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
