Given f'(x)=(2-x)(6-x), determine the intervals on which f(x) is increasing or decreasing
Options:
1) Decreasing (-∞,2); increasing on (6,∞)
2) Decreasing (2,6); increasing on (-∞,2)U(6,∞)
3) Decreasing (-∞,2)U(6,∞); increasing on (2,6)
4) Increasing (-∞,-2)U(-6,∞); increasing on (-2,-6)
decreasing between those of course
To determine the intervals on which f(x) is increasing or decreasing, we can analyze the sign of the derivative f'(x).
The derivative f'(x) is given by (2-x)(6-x).
To find the critical points, we set f'(x) equal to zero and solve for x:
(2-x)(6-x) = 0
Setting each factor equal to zero gives us:
2-x = 0 --> x = 2
6-x = 0 --> x = 6
We have two critical points at x = 2 and x = 6.
Now, we can analyze the sign of f'(x) on different intervals:
For x < 2, both factors (2-x) and (6-x) are positive, so f'(x) > 0.
This means that the function is increasing on the interval (-∞, 2).
For 2 < x < 6, the factor (2-x) is negative and the factor (6-x) is positive, so f'(x) < 0.
This means that the function is decreasing on the interval (2, 6).
For x > 6, both factors (2-x) and (6-x) are negative, so f'(x) > 0.
This means that the function is increasing on the interval (6, ∞).
Therefore, the correct option is:
3) Decreasing (-∞,2)U(6,∞); increasing on (2,6)
To determine the intervals on which the function f(x) is increasing or decreasing, we need to find the critical points. Critical points occur where the derivative f'(x) is either zero or undefined.
First, let's find the critical points by setting f'(x) equal to zero:
(2 - x)(6 - x) = 0
This equation will be satisfied when either (2 - x) = 0 or (6 - x) = 0.
1) When (2 - x) = 0, solving for x gives x = 2.
2) When (6 - x) = 0, solving for x gives x = 6.
So, we have two critical points: x = 2 and x = 6.
To determine the intervals of increase and decrease, we can test the sign of the derivative (f'(x)) on each interval. If the derivative is positive, f(x) is increasing. If the derivative is negative, f(x) is decreasing.
Now, let's test the sign of f'(x) on different intervals:
For x < 2, pick a value less than 2 (e.g., x = 0):
f'(0) = (2 - 0)(6 - 0) = 12 > 0
So, f'(x) is positive for x < 2, which means f(x) is increasing on the interval (-∞, 2).
For 2 < x < 6, pick a value between 2 and 6 (e.g., x = 4):
f'(4) = (2 - 4)(6 - 4) = -4 < 0
So, f'(x) is negative for 2 < x < 6, which means f(x) is decreasing on the interval (2, 6).
For x > 6, pick a value greater than 6 (e.g., x = 8):
f'(8) = (2 - 8)(6 - 8) = 12 > 0
So, f'(x) is positive for x > 6, which means f(x) is increasing on the interval (6, ∞).
Therefore, the correct option is:
1) Decreasing (-∞,2); increasing on (6,∞)
where is first derivative, f' positive ?
f'(x)=(2-x)(6-x) that is a parabola,
coef of x^2 is + so the vertex is at the bottom
zero when x = 2 and when x = 6 which is
so when x < 2 and when x > 6
so to the left of x = 2 and to the right of x = 6