How do ik when I can combine answers and not?
#trigonometry
steady sorrel
pale remnant
How do ik when I can combine answers and not?
That's not correct. You can't combine answers like that here.
Usually you can see when to combine them if you mark them on a unit circle.
steady sorrel
I have the unit circle with set numbers but idk what to do when I get weird angles i can't directly know where they are
Is there another way?
pale remnant
Hm, good question... Not sure if there's a general way. Such a thing isn't done too often, really.
steady sorrel
How do you know I can't combind the answers here?
pale remnant
How do you know I can't combind the answers here?
Well, here I visualized them on a unit circle. They are not equidistant from each other, so they can't be combined in the way you wrote.
steady sorrel
I see now
Big thanks!
I have a few more questions
Do I need to make a new post?
pale remnant
Nah, you can ask here.
pale remnant
Alright. So, what did you do?
steady sorrel
I found two extreme points solving the derivative x=1 and x=-2
But in the answer
It says ut has one for x=-3 and x=3 too
Why
pale remnant
It says ut has one for x=-3 and x=3 too
Because when you analyze the function on a closed interval, the endpoints must also be considired.
So, you need to compare not only the values at critical points, but also the values of the endpoints.
So, in this case you need to compare f(-3), f(-2), f(1) and f(3).
steady sorrel
Are the ends always critical points? Or are they just that if they're bigger/smaller than everything else in y values?
pale remnant
Are the ends always critical points? Or are they just that if they're bigger/sma...
No, the ends aren't critical points by themselves, but they still need to be compared to the values at cricial points, if there are any.
For example, consider f(x) = 2 - x for 0 ≤ x ≤ 1.
Clearly, f(x) has no critical points, but as it is continuous and defined on a closed interval, then by Weierstrass's theorem it must reach its minimum and maximum values on that interval. And it does, just at the endpoints: the maximum is f(0) = 2 and the minimum is f(1) = 1.
steady sorrel
I see
Just feels weird how something with a biggest/smallest value where its defined
Can be called the same as
When the derivate is 0
Or does it have a different name?
pale remnant
Well, the critical points are the points at which the derivative is zero or doesn't exist.
steady sorrel
What you mean by doesn't exist?
pale remnant
What you mean by doesn't exist?
Consider f(x) = |x|. f'(0) doesn't exist.
steady sorrel
Yes
pale remnant
But x = 0 is obviously the minimum point.
So, we also need to consider the points where the derivative doesn't exist (but where the function itself is still defined).
pale remnant
You're welcome!
What else did you need?
steady sorrel
So intergals
When x^-1 becomes ln|x|
Why is it ln|x| and not ln(x)
What's even the difference
Is it only positive nunbers
Or...
pale remnant
Why is it ln|x| and not ln(x)
Because ln(x) is defined for x > 0, but 1/x is defined for x ≠ 0.
We don't want to lose half of the domain when integrating, so the antiderivative of 1/x is ln(|x|).
pale remnant
I assume you want to find the area of that region between the parabolas?
steady sorrel
Yes and I did but
I don't understand why it's the yellow area minus the red
How do you visualise this
pale remnant
No, it's not like that.
The function is bounded above by 8 - x^2/8 and below by x^2/4 + 2, so you need to integrate (8 - x^2/8) - (x^2/4 + 2).
To find the limits of integration, you need to find out where the parabolas intersect. To do that, solve 8 - x^2/8 = x^2/4 + 2.
pale remnant
Yeah, looks good.
steady sorrel
I first make a new function, the top area (yellow) minus the one below (red)
And when I do that
I first have the yellow area
And then take away everything that's not touching both at the same time
I'm confused
Visually
If I had two apples
No
steady sorrel
But I don't understand why I took the yellow minus the red function
Visually in the graph that confuses me
pale remnant
You take the function that bounds the region above and subtract the one below.
So, in this case, orange minus red.
steady sorrel
So it's always the one above minus the one beneath?
pale remnant
Yes.
