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To determine the intervals of increasing and decreasing for the function [tex]\( f(x) = x^2 - 12x + 35 \)[/tex], we follow these steps:
1. Find the first derivative of the function:
The first derivative, [tex]\( f'(x) \)[/tex], represents the rate of change of the function and can be computed as follows:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 12x + 35) = 2x - 12 \][/tex]
2. Determine the critical points:
Critical points occur where the first derivative is equal to zero or where it does not exist. In this case, set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 12 = 0 \implies x = 6 \][/tex]
So, the critical point is [tex]\( x = 6 \)[/tex].
3. Use the second derivative to determine concavity:
The second derivative, [tex]\( f''(x) \)[/tex], helps to determine if the function is concave up or concave down at the critical points:
[tex]\[ f''(x) = \frac{d}{dx}(2x - 12) = 2 \][/tex]
Since [tex]\( f''(x) = 2 \)[/tex] is positive, the function is concave up everywhere.
4. Determine the intervals of increase and decrease based on the critical points and concavity:
- Since the second derivative is positive, the function is concave up, meaning the function has a minimum at the critical point [tex]\( x = 6 \)[/tex].
- For [tex]\( x < 6 \)[/tex] (interval to the left of the critical point), based on the concavity, the function is decreasing.
- For [tex]\( x > 6 \)[/tex] (interval to the right of the critical point), based on the concavity, the function is increasing.
Therefore, the intervals of increasing and decreasing for the function [tex]\( f(x) = x^2 - 12x + 35 \)[/tex] are:
- The function is increasing on the interval: [tex]\( (6, \infty) \)[/tex].
- The function is decreasing on the interval: [tex]\( (-\infty, 6) \)[/tex].
1. Find the first derivative of the function:
The first derivative, [tex]\( f'(x) \)[/tex], represents the rate of change of the function and can be computed as follows:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 12x + 35) = 2x - 12 \][/tex]
2. Determine the critical points:
Critical points occur where the first derivative is equal to zero or where it does not exist. In this case, set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 12 = 0 \implies x = 6 \][/tex]
So, the critical point is [tex]\( x = 6 \)[/tex].
3. Use the second derivative to determine concavity:
The second derivative, [tex]\( f''(x) \)[/tex], helps to determine if the function is concave up or concave down at the critical points:
[tex]\[ f''(x) = \frac{d}{dx}(2x - 12) = 2 \][/tex]
Since [tex]\( f''(x) = 2 \)[/tex] is positive, the function is concave up everywhere.
4. Determine the intervals of increase and decrease based on the critical points and concavity:
- Since the second derivative is positive, the function is concave up, meaning the function has a minimum at the critical point [tex]\( x = 6 \)[/tex].
- For [tex]\( x < 6 \)[/tex] (interval to the left of the critical point), based on the concavity, the function is decreasing.
- For [tex]\( x > 6 \)[/tex] (interval to the right of the critical point), based on the concavity, the function is increasing.
Therefore, the intervals of increasing and decreasing for the function [tex]\( f(x) = x^2 - 12x + 35 \)[/tex] are:
- The function is increasing on the interval: [tex]\( (6, \infty) \)[/tex].
- The function is decreasing on the interval: [tex]\( (-\infty, 6) \)[/tex].
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