Given f'(x)=(x-4)(6-2x) find the x-coordinate for the relative minimum of the graph f(x).
OPTIONS:
8
6
3
None of these
I think it is 3, but I am also conflicted to say none of these because I graphed the function and ant see a minimum.
*cant
To find the x-coordinate for the relative minimum of the graph of f(x), we need to analyze the sign of the derivative f'(x).
The derivative is given as f'(x) = (x-4)(6-2x).
To determine the sign of the derivative, we can examine the factors individually.
For the factor (x-4), the sign changes from negative to positive as x increases from 4 to infinity.
For the factor (6-2x), the sign changes from positive to negative as x increases from -infinity to 3, and then changes from negative to positive as x increases from 3 to infinity.
Now, let's consider the signs of both factors together.
When (x-4) and (6-2x) have the same sign, their product is positive.
- When x < 4, both factors are negative, so (x-4)(6-2x) is positive.
- When 3 < x < 6, the first factor (x-4) is negative, and the second factor (6-2x) is positive, so (x-4)(6-2x) is negative.
- When x > 6, both factors are positive, so (x-4)(6-2x) is positive.
From this analysis, we can see that there is a change in sign at x = 3, indicating a possible relative extremum. However, to determine whether it is a minimum or maximum, we need to consider the behavior of the function.
Since you mentioned that you graphed the function and did not observe a minimum, it is likely that there is no relative minimum. Therefore, the correct answer is "None of these."
To find the x-coordinate for the relative minimum of the graph of f(x), you need to find the critical points of the function f(x) and then check the second derivative to determine whether each critical point is a relative minimum or maximum.
The critical points occur when the derivative of the function, f'(x), equals zero or does not exist. In this case, we are given f'(x) = (x-4)(6-2x). To find the critical points, we set f'(x) equal to zero and solve for x:
(x-4)(6-2x) = 0
To solve this equation, we can use the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero. So, we can set each factor equal to zero and solve for x:
x - 4 = 0 => x = 4
6 - 2x = 0 => x = 3
Therefore, the critical points are x = 4 and x = 3.
To determine whether each critical point is a relative minimum or maximum, we need to check the second derivative, f''(x). However, the second derivative is not given in the question. So, we cannot determine whether the critical points correspond to a relative minimum or maximum without further information.
Therefore, the correct answer is "None of these," as we do not have enough information to determine the x-coordinate for the relative minimum.
clearly it is either 3 or 4.
Since 4 is not a choice, 3 seems logical.
Now, knowing what you do about cubic curves, and that this one comes down from the left, its first turnaround will be a minimum. So, it will be at x=3.
Or, note that f" = -4x+14.
f"(3) = 2 > 0, so f(3) is a minimum.
I can't see how you could have graphed it and not seen a minimum.