Interesting piece-wise problem

[mathgal]

I hate a piece-wise function

f(x)= [ 1 if 0 </eq x </1
[ 0 if 1 < x < 2
[ -2x+4 if x >/eq 2

If I let F(x)= integral (from 0->x) f(t) dt, how can i find a formula for F(x) without using integrals? I think it may have something to do with the 2nd definition of the Fundamental Theorem of Calculus, but I'm not sure.

 

[Mark44]

I hate a piece-wise function

f(x)= [ 1 if 0 </eq x </1
[ 0 if 1 < x < 2
[ -2x+4 if x >/eq 2

Use <= instead of </eq, and >= instead of >/eq.

If I let F(x)= integral (from 0->x) f(t) dt, how can i find a formula for F(x) without using integrals? I think it may have something to do with the 2nd definition of the Fundamental Theorem of Calculus, but I'm not sure.

Have you graphed this function? After you do this, you should be able to come up with a formula for F(x), using nothing more sophisticated than geometry.

 

[mathgal]

Should I be using limits? I actually did graph it. I know what the F(x) should be, I'm just not sure how to rigorously show it using a theorem, formula, etc

 

[Mark44]

No, you don't need limits. This is mostly a geometry problem. Your formula for F(x) will be defined in a piecewise fashion.
To get you started,
F(x) = x, 0 <= x < 1

You will need two more definitions, one for each of the two additiona intervals. The formula for F(x) for 1 <= x < 2 is very easy.

 

[mathgal]

Is this correct? I'm worried the last one should be more complicated?

F(x) = x, 0 <= x <= 1
= 0, 1 < x < 2
= -x^2+4x, x >= 2.

 

[mathgal]

Well it should actually be F(x)=-x^2+4x-4 for x>= 2 right?

 

[The Chaz]

I hate a piece-wise function

Just one? I hate ALL piece-wise functions!

 

[mathgal]

Also, is there a way for me a find a formula for F'? How do I first find where F is differentiable?

 

[Mark44]

Is this correct? I'm worried the last one should be more complicated?

F(x) = x, 0 <= x <= 1
= 0, 1 < x < 2
= -x^2+4x, x >= 2.

The second and third parts are wrong. Remember that F(x) is defined as
$$\int_0^x f(t) dt$$

So, in particular, F(1.5) $\neq$ 0, as you have, and F(2) $\neq$ 4, as you have. F(2) should actually be equal to 1.

 

[Mark44]

Just one? I hate ALL piece-wise functions!

I divide functions into three categories:

• the ones I hate
• the ones I love
• the ones that I feel neutral about

 

[mathgal]

so should F(x)=x also if 1<x<2?
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 0

is that correct?

 

[mathgal]

Actually shouldn't F(x)= 1, 0<=x<=1, F(x)=1, 1<x<2, and F(x)=-x^2+4x-3, x>=2

 

[Mark44]

Edit: This refers to post 11.
No, that's not it either. For example, what is F(1.5)? It's not 1.5.

int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 0

I'm going to translate part of the line above into actual mathematics.
$$F(2) = \int_0^1 1~dx + \int_1^2 0~dx$$
The value of the first integral is not x, and x + 0 is not 0.

 

[Mark44]

Edit:Changed my answer.

Actually shouldn't F(x)= 1, 0<=x<=1, F(x)=1, 1<x<2, and F(x)=-x^2+4x-3, x>=2

Well, you're close. The first part should be F(x) = x, 0 <= x <= 1.

 

[mathgal]

But we're evaluating it from (0,1)... So once we integrate and get x|, then we have 1-0=1. This isn't correct?

 

[mathgal]

Oh and in Post #11, I meant to say:
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

 

[mathgal]

Oh and in Post #11, I meant to say:
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

 

[Mark44]

Oh and in Post #11, I meant to say:
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

A lot of this is gibberish. F(x) is not the integral from 1 to 2 and it's not the integral from 0 to 2, and it's not x, which is not 1.

F(x) needs to be defined in piecewise fashion.

 

[mathgal]

okay well thanks, anyways

 

[Mark44]

I'm not trying to discourage you (although I might be succeeding), but I am trying to get you to write things that make sense.

 

[mathgal]

F(x)= x if 0 <= x <= 1.

Then, for 1< x <2, we're taking the integral of 0, so I should just get a constant?
I just don't know how I am supposed to write that part of the third part whre x>=2

 

[Mark44]

$$F(x) = \int_0^x f(t) dt$$
If 0 <= x <= 1, F(x) = x
If 1 < x <= 2, F(x) = 1 + 0 = 1. The 1 comes from the rectangle above the interval [0, 1].
For the 2nd and 3rd intervals you have to add the contribution(s) from the previous interval(s).

It's really important to have a good graph of the function F, showing the three parts. The first part is a line with slope 1 between (0, 0) and (1, 1). the second part is a horizontal line between (1, 1) and (2, 1). The third part is the right half of a parabola that opens downward and whose vertex is at (2, 1). At the point where the parabola crosses the x-axis, F of that x is zero. From that point onward, F(x) < 0.

 

[mathgal]

I think I actually understood that better than I thought and was just getting confused by my own writing. Thank you so much. I know you've already helped alot, but do you have any ideas about how I would then find a formula for F'(x) wherever F is differentiable? My test is tomorrow, that is the only reason I'm trying to learn this stuff now.

 

[Mark44]

The graph of F is continuous for x >= 0. The only points where it is not differentiable are at x = 1 and x = 2. The formula for F'(x) will have three parts.

For 0 < x < 1, F(x) = x, so F'(x) = 1. Do the same for the other two parts, but keep in mind that F isn't differentiable at x = 1 or x = 2 (or at x = 0). Here's where you might start thinking about the Fund. Theorem of Calculus.

 

[mathgal]

But doesn't the Fundamental Theorem of Calculus tell us that since our F is continuous everywhere, that it should be differentiable there? How do I show that x=1,2 are not differentiable? I know if the limit from the left and right are equal, it is continuous at that point.

 

[Mark44]

But doesn't the Fundamental Theorem of Calculus tell us that since our F is continuous everywhere, that it should be differentiable there?

It can't say that, since it isn't true.
For example, the graph of f(x) = |x| is continuous everywhere, but not differentiable at x = 0. The derivative does exist, however, for x < 0 (f'(x) = -1) and for x > 0 (f'(x) = 1).

How do I show that x=1,2 are not differentiable? I know if the limit from the left and right are equal, it is continuous at that point.

Don't say that "x = 1,2 are not differentiable." What you should say is that F is not differentiable at x = 1 and x = 2.

Also, I think you might be confusing the idea of continuity with differentiability. Intuitively, a continuous function is one whose graph can be drawn without taking your pen from the paper - no jumps, no holes. Differentiability puts more constraints on the graph - there can't be any corners (cusps) either.

The graph of F(x) in your problem is continuous for all x >= 0, but has "corners" at x = 1 and x = 2 that keep it from being differentiable at those points. F is differentiable at interior points in its domain other than those.

 

[mathgal]

So for my formula for F', is there something specific I have to write for x=1 and x=2? How would I do that since my intervals are 0<= x <= 1 and 1<x<2 and x>=2 where there's ones and twos overlapping?

 

[Mark44]

Your intervals for F'(x) won't include 0, 1, or 2, that's all. IOW, you'll have one formula for 0 < x < 1, another for 1 < x < 2, and another for x > 2.

 

[mathgal]

So I have:

F'(x)= 1 if 0<x<1
0 if 1<x<2
-2x+4 if x>2

Like that?

 

[Mark44]

Bingo! And the only difference between F'(x) and f(x) in your first post is the endpoints of the intervals.