I was given this problem at school.
f(x) = |x-4|+7 on the interval 1 < x < 6
I don't understand what the |x-4| means. Are those supposed to be the same as brackets?
If so, is f(x) = (x-4)+7
Then f '(x) = 1?
Thank you.
I was given this problem at school.
f(x) = |x-4|+7 on the interval 1 < x < 6
I don't understand what the |x-4| means. Are those supposed to be the same as brackets?
If so, is f(x) = (x-4)+7
Then f '(x) = 1?
Thank you.
The vertical lines are absolute value symbols. Basically, |x| measures how far x is from zero. You can also think of the absolute value signs as erasing the sign information. That is, |x| is always non-negative. The derivative of |x| exists everywhere except at x = 0. So the derivative of |x-4| exists everywhere except at x = 4. Since that's in your interval of interest, that'll be a critical point.
Can you finish from here?
I am still confused about the given function though
I know the derivative of 7 is going to be 0 because it is a constant, but how do I get the derivative of |x-4|
You said that it exists everywhere except at x= 4. That much I understand, but I don't know what to do with that information.
Just take the derivative of the expression TheEmptySet has for you. It'll be a piece-wise defined function. You can see where the derivative doesn't exist, and when it is zero (which is actually never). Then apply your usual maximization/minimization algorithm.
Okay guys, I gotta be honest. I'm almost completely lost here.
So far, they have taught me how to calculate the maximum and minimum points of a function. I also just learned how apply the product function when multiplying the derivatives. However, they just threw me off with this |x| absolute value stuff. I could wait until my teacher explains this in class but I would prefer to understand this beforehand.
Would you mind explaining a bit further?
Thanks in advance.
Ok. The method for finding the extremes of a continuous function on a closed interval is as follows:
1. Find all points where the derivative is either zero or does not exist. All of these points are called the critical points.
2. Evaluate the function at the critical points and the endpoints of the closed interval.
3. Then the maximum value found in step 2 is the maximum, and the minimum value found in step 2 is the minimum.
1. So let's look at your function f. Taking the derivative of TheEmptySet's expression in post # 3 yields the following:
The derivative does not exist at x = 4, because the function has a corner there, and the limit of the slopes of the secant lines from the right does not equal the limit of the slopes of the secant lines from the left. So, from this derivative I've shown you, you can see that it's never zero, but it is undefined at x = 4. Hence, the only critical point is at x = 4.
2. I have a question: are you absolutely sure that you're supposed to maximize and minimize f(x) on the open interval (1,6)? That's what you've got there: 1 < x < 6. This function doesn't have a maximum value on that interval, though it does have a minimum value. If you meant the closed interval [1,6], with 1 <= x <= 6, then the function, being continuous on a closed interval, will have a maximum. In that case, you'd evaluate f at the critical point x = 4, and at the endpoints x = 1, and x = 6. What values of the function do you get for those three values?
I can't really tell because the way it is written in the lab is as 1 £ x £ 6.
I am going to assume that it is 1 <= x <= 6 because it really makes more sense.
So let's see what I got here.
f'(x) = 1 when x > 4
f'(x) = -1 when x < 4
f(1) = 1 + 3 = 4 if x => 4
f(1) = -1 + 11 = 10 if x <=4
f(4) = 4 + 3 = 7 if x =>4
f(4) = -4 + 11 = 7 if x<= 4
f(6) = 6 + 3 = 9 if x => 4
f(6) = -6 + 11 = 5 if x <= 4
So my points would be (1, 4), (1, 10), (4, 7), (4, 7), (6, 9), (6, 5)
I have no idea if what I'm doing is correct at this point.
Also, would you mind explaining me what is it that TheEmptySet did on post #3? I can see where the 3 and the 11 come from but I don't understand the process
Some of your computations are extraneous. Here's the way you handle piecewise defined functions like in Post # 3.
Let's say you want to evaluate f(1). Well, in this case, x = 1. Now you go to your piecewise defined function, and find out which region contains the number 1. You have two possibilities: is 1 >= 4, or is 1 < 4? Clearly the latter, so you must plug in 1 into the bottom expression. That is, f(1) = -1 + 11 = 10. f(1) does NOT EQUAL 4. Why not? Because the portion of the function corresponding to x + 3 only applies when x is greater than or equal to 4.
So, based on this analysis, can you give me, with certainty in your mind, the values of f(4) and f(6)?
I will give it a try,
Now I want to evaluate f(4) and f(6), so x = 4 and x = 6. That means that I need to find the region which contains the numbers 4 and 6 in my piecewise function. That region is x >= 4.
Therefore;
f(4) = 4 + 3 = 7
f(6) = 6 + 3 = 9
Can I say that my critical points are (4, 7) and (6, 9)? How can I find the max and the min from these points?
Finding the max and min of these points is quite simple. You've already done all the hard work! Your three points at which the global max and min MIGHT occur are the following:
(1,10), (4,7), and (6,9). For which of these points is the y-value the greatest? That's your max. For which of these points is the y-value the least? That's your min.