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### 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]
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