Given f '(x) = (2 - x)(6 - x), determine the intervals on which f(x) is increasing or decreasing
f'(x) is negative between the roots of 2,6 and positive otherwise.
So, f is increasing on (2,6) and decreasing elsewhere.
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 need to solve the equation f'(x) = 0.
(2 - x)(6 - x) = 0
Setting each factor equal to zero gives us the critical values:
2 - x = 0 or 6 - x = 0
Solving these equations, we find:
x = 2 or x = 6
Now we have the critical points x = 2 and x = 6. These points divide the number line into three regions:
Region 1: x < 2
Region 2: 2 < x < 6
Region 3: x > 6
Now, let's determine the sign of the derivative f'(x) in each of these regions.
For Region 1, we can choose a test value x = 0:
f'(0) = (2 - 0)(6 - 0) = (2)(6) = 12, which is positive.
Therefore, f(x) is increasing in Region 1.
For Region 2, we can choose a test value x = 4:
f'(4) = (2 - 4)(6 - 4) = (-2)(2) = -4, which is negative.
Therefore, f(x) is decreasing in Region 2.
For Region 3, we can choose a test value x = 7:
f'(7) = (2 - 7)(6 - 7) = (-5)(-1) = 5, which is positive.
Therefore, f(x) is increasing in Region 3.
By analyzing the signs of the derivative f'(x) in each region, we can conclude:
- f(x) is increasing for x < 2 and x > 6.
- f(x) is decreasing for 2 < x < 6.
To determine the intervals on which f(x) is increasing or decreasing, we need to analyze the sign of the derivative function f'(x).
Let's start by finding the critical points of f(x), which are the values of x where f'(x) is equal to zero or undefined.
f'(x) = (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 two possible critical points:
2 - x = 0 or 6 - x = 0
Solving each equation for x, we find:
x = 2 or x = 6
So, the critical points are x = 2 and x = 6.
Next, we create a number line and mark the critical points on it:
----------------------
2 6
Now, we select test points from each of the intervals created by the critical points and evaluate f'(x) at these test points. This will determine the sign of f'(x) in each interval.
For example:
- Let's choose a test point less than 2, say x = 0.
- Evaluating f'(x) = (2 - x)(6 - x) at x = 0, we get f'(0) = (2 - 0)(6 - 0) = 12.
- Since f'(0) = 12, which is positive, we can conclude that f'(x) is positive for values of x less than 2.
Similarly, we can choose test points in the intervals (2, 6) and (6, infinity) to evaluate f'(x) and determine the sign.
Using these test points, we can construct a sign chart for f'(x):
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Interval | Test Point | f'(x) = (2 - x)(6 - x) | Sign of f'(x)
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(-infinity, 2) | x = 0 | (2 - 0)(6 - 0) = 12 | +
(2, 6) | x = 3 | (2 - 3)(6 - 3) = -3 | -
(6, infinity) | x = 7 | (2 - 7)(6 - 7) = 5 | +
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Now, we can determine the intervals where f(x) is increasing or decreasing based on the sign of f'(x):
- f(x) is increasing on the intervals (-infinity, 2) and (6, infinity) since f'(x) is positive in these intervals.
- f(x) is decreasing on the interval (2, 6) since f'(x) is negative in this interval.
Therefore, f(x) is increasing on the intervals (-infinity, 2) and (6, infinity), and f(x) is decreasing on the interval (2, 6).
d/dx [ (2-x)(6-x) ] = slope of f'(x), where is it + ?
= (2-x)(-1) + (6-x)(-1)
= -2+x -6 + x = 2 x-8
that is + if x >4
and it is - if x<4