Hint: The way you solve this problem determines whether you belong on Veeky Forums or not

too trivial to even answer

A
f(x) = -|x-1|+1 = 0
x=0 and x=2

b
Intersections are simply where f(x) = F(x). That is all.

This is the only correct response so far and, of course, will remain so. Perhaps op thought the lower case f(t) would invite responders to substitute f(x)? I dont know but it's a silly question.

this
also
/thread

The graphs correspond to the functions right?
ok how about these
a) (3.4, 0)
b) (0,0) and there is no way to figure out the second point from the given information

>there is no way to figure out the second point from the given information
except by integrating...

>trivial
>every actual answer is wrong

You failed, and you didn't even get a right answer.

I would solve this visually without doing any math whatsoever, deal with it. I'm an engineer btw.

Do you know what that function is?

I don't really know shit desu. Is F(x) supposed to be the integral of the fourier transform of f(x)?

a) at x=0 and x=3.4
b) at x=0 and x=1.24

ez visually

It asks for the roots of F(x) not f(x). But let's assume the blue curve is not correctly drawn.
F(x) = integral f(t) dt from t =0 to t= x
F(x) = F(t) from t =0 to t= x
F(x) = F(x) - F(0)
F(0) =0 so
F(x) = F(x)
So it doesn't tell us anything.

it's the integral of the absolute value form given

just the intersection of a parabola with a line

I don't even know calculus lol

a) (0;0)
b) (1.25; 0,75)

>[math]F(x)=\int^x_0 f(t)dt[/math]
I'm assuming that's supposed to be [math]F(x)=\int^x_0 f(x)dx[/math]?

The graph confirms it considering that the gradient of F(x) is 0 when the value of f(x) is 0 so dF/dx = f(x)

Ok since I don't know calculus
a) 0,0 is trivial next is 2 + sqr(2) from the definition of integral (triangle above x axis has area of 1 and the next triangle is made by a linear function with a = 1)
b) I get 2 - x = 0.5 [value of area from integral up until x = 1) + integral of x - 1 transformed by vector [-1,0] so I get equation 2 -x = 0.5 + (x - 1) - (x-1)^2 x 1/2
From this it's just roots of second power equation so I get 3 - srt(3) and 3 + srt(3)

Brainlet/10

the OP has typed it correctly

wrong

retard

>what is a dummy variable

I think this thread confirms Veeky Forums is filled for the most part with mouth breathers who can't even deal with high school level calculus.

well you got there eventually

Take your pedophile cartoons back to . Fucking degenerate.

Right. I regularly fuck up maths and practically flunked uni but here goes:

a) [math](0,0)[/math] and [math](2+\sqrt{6},0)[/math]

b) [math](0,0)[/math], [math](3+\sqrt{3},-1-\sqrt{3})[/math] and [math](3-\sqrt{3},-1+\sqrt{3})[/math]

>I think this thread confirms Veeky Forums is filled for the most part with mouth breathers who can't even deal with high school level calculus
Then just kill me, sempai. I only came to Veeky Forums from /trash/ to ask whether the remainder of any number, after dividing by zero, is zero.

I mean, you want it in some way that isn't solving the integral? Thebn just equate areas for the first one. You now from 0 to 2 its positive and the area there is 1 so just 1=f(x)/(x-2). Then for the next one after equating the two func take the derivative of both sides and solve for x.

>I got a) wrong
Fine. [math](0,0)[/math] and [math](2+sqrt{2},0)[math].
I fucked up my maths and added 2 in the square root rather than took 2.

I don't care how it's solved, I'm interested in seeing what solution approaches people come up with. Solve everything geometrically without the use of integrals or derivatives is just one example.

If |x-1|=1, then the distance between x and 1 is 1. This gives x=2 or
x=0, obviously. So (0,0) and (2,0) is the answer to b.

If x1.

Read part a again

A: x=0 or 1 - 1/2*(x-2)^2 = 0, x>2
x = 0 or x =sqrt(2) + 2
B: -|x-1|+1 = ∫-|x-1|+1dx
= ∫xdx (from 0 to 1)+∫2-xdx (from 1 to x)
= 1/2 -x^2/2+2x-3/2
-x+2 = -1 -x^2/2+2x for x>=1
0=x^2-6x+6
0=(x-3)^2-3
0=(x-3+sqrt(3))(x-3-sqrt(3))
x = 3-sqrt(3) or x = 3+sqrt(3)
final solution x = 0, 3-sqrt(3), 3+sqrt(3)

How did I do?

I forgot to give the exact points but you can just sub the x values back into f(x). Also line 4 should really state that x>=1.

lmao

but but Veeky Forums is full of geniuses! Had you posed this as a question rather than a challenge, everyone would be sitting here laughing and calling you a brainlet

>you should word your dumb question better, is b) asking for the intersections of F(x) & f(x) or f(x) and the x axis?
a. (0,0) and ([math]\sqrt{a}[/math]+2,0)
b. (0,0) and (2,0)
What did I win?

fucking latex
*([math]\sqrt{3} +2[/math],0)

Is this bait?
That's clearly not what b is asking. It wants the intersection of F(x) and f(x)

I'd say you've done well apart from forgotting to include the y values and (0,0) for b) as well.
But I'm not OP.

I did include x = 0 for b, but yeah I forgot to give y values.

[eqn]\int-|x-1|+1[/eqn]
[eqn]-\frac{\left(\left|x-1\right|-2\right)x-\left|x-1\right|}{2}-\frac{1}{2}[/eqn]
[eqn]\frac{1}{2}\left(\left(x-1\right)^2\operatorname{sign}\left(1-x\right)+2x-1\right)[/eqn]

to lazy to step it out but you get the idea

>I did include x = 0 for b
Ah, drat. Half two in the morning isn't good for making observations.

Everyone defaults to calculus when they see an integral. The function f(x) is two lines:

y=x for x=1

Just use the formula for the area of triangle:

a) 1 + 1/2(X-2)^2 = 0 (the height = base = x-2)
b) 1 - 1/2(2-x)^2 = -x +2 (1 minus the base of small triangle to the left of the intersection point, equal to the equation of the line for x>=1)

I'll let you do the math but:

a) 2 + sqrt(2)
b) 3 +- sqrt(3)

Note 0 is the trivial answer so not included above.

>had to use mathematica to solve a basic calc problem

I seriously hope you pirated it.

don't you need a third equation relating t to x or y in order to be able to evaluate the second equation?

do you need to know anything relating k to n in order to be able to evaluate [math]\sum_{k=0}^nk^2[/math]?

f(t) = -|t-1|+1

no. but that is a summation. Consider f(t)=2x^3+e^5x. from what information OP has provided I don't know that it doesn't. in that case your a cunt.

maybe, funny coincidence thats also what f(x) equals.

Right, I think I remember now.
For the second equation, what's actually going on is [math]F(x)=\int^{t=x}_{t=0}f(t)dt[/math]
I'm guessing it's usually assumed... but I forgot about it myself despite working it out correctly.
I have no idea what you're going on about with n or k.

Are you just pretending to be retarded or do you really not understand what a bound variable is?
[eqn]\int_a^bx^2 dx=\int_a^b y^2 dy=\int_a^b t^2dt[/eqn]
It doesn't matter what you call your dummy variable, the expression means exactly the same.

OP never made it clear that f(t)=f(x). I would never feel comfortable assuming it did.

...

a) Solving by properties of triangles + trivial solution:
(x,y) = (0,0) and (2 + sqrt(2), 0)

b) Solving by integration over negative sloped region + trivial solution:
(x,y) = (0,0), (3 - sqrt(3), sqrt(3) - 1), and (3 + sqrt(3), -1 - sqrt(3))

Are you reading the posts you're replying to?

On the off-chance that you aren't pretending to be retarded:

f(x) and f(t) are essentially the same function, just with t swapped in for x. They are both called "f" so that is a given. If it was called g(t) that would be different.

Accepting this, you are right that x does not necessarily equal t just yet. However, it does once you apply the equation in part b -- this defines that F(x) is equal to the definite integral of f(x) from 0 to any arbitrary x. Solve the integral yourself using the dummy variable t and plug in x if you don't believe me. There is no ambiguity here.

You're all idiots. Thanks for the cheap laugh.

----------------------------idiot-line----------------------------
^everyone above this line is an idiot

heh, jokes on you

Come back when you're old enough to be here

it's two second degree polynomials stitched together at x=1. the graph passes through (1,0.5) because 0,5 is the area of the triangle with corners (0,0) (0,1),(1,1). So we want to find some intersections for the parabola passing through (1,0.5) and having maximum at (2,1). Now use the information to find a formula for the parabola and you should be fine finding the intersections.

no u X^D

e.z.

you're missing 2 points you brainlet, kys

strictly speaking F(x) doesnt intersect the x-axis at (0,0), it only makes contact. intersect means to divide into two sections.

kys illiterate tard, and learn to derive your self esteem from something other than math problems where you arbitrarily reject otherwise valid answers.

Still missing one point :^)

a) 0, 2+sqrt(2)
b) 0,3-sqrt(3),3+sqrt(3)

>The way you solve this problem determines whether you belong on Veeky Forums or not.
Nice working faggot. Any retard can plug the problem into a calculator.

Bet you feel dumb now.

What was that?

this tbqh
stay mad brainlet

I don't think you understand. I wasn't the original poster, but you're regardless of whether or not (0,0) is an answer your forgetting a third intercept for part b

In mathematical terminology intersection is more general and means what two or more sets, geometric objects, etc. share in common.

It doesn't really matter whether it "passes" through or "makes contact", they still share a point in common and therefore intersect.

If it helps your intuition, you can think of the x-axis as being divided into two parts before and after the "contact" point.

This is mostly semantics. Anyway, kys.

I'm not that guy, I'm calling him a brainlet

I'm not the one doubleposting :^)
Oh, my b

>he didn't solve part a) in an easier way by using triangles
GET OFF MY BOARD

The answer is; I don't give a fuck.
The """scientific""" and """intellectual""" discussion on this board is shit tier at best.

The easier way would be to use Mathematica

2bh part a) would be solved faster with just your brain

This guy gets it

This is why I fucking hate this board ..everyone thinks they're so damn smart n shit and how other ppl are insignificant. Respect average ppl cuz without them u wouldn't be ABOVE average.

Uninteresting problems get uninteresting solutions.

interesting thread, i thought change of variable was one of the most basic things taught in calculus but i guess not

you have to go back

he gets it, but he's not explicitly explaining steps. good student, poor teacher.

...

OP for
Thanks for the feedback. When I was in graduate school I once invoked the Fundamental Theorem of Abelian Groups in a proof. The prof said "You just hit a nail with a sledgehammer" meaning it was technically accurate but the aspect I needed was a fairly trivial two line independent proof. Best prof I ever had.

If I had a white board to draw on, it would become obvious. Show some initiative and draw a diagram on paper. I gave you everything you need to solve the problem. It's about triangles for which the basic area (i.e. integral) = 1/2 base x height. I am not going to hold your hand any more than that.

You forgot another point, there is an intersection further to the right.
A) 2+sqrt(2) and 0
B) 3 - sqrt(3) and 3+sqrt(3)

The solution contains a trivial answer 0 for both problems

a)
Let F(x) = 0
Suppose sum f()(0,2) - sum f() (2,x) = 0
So sum f() (0,2) = 1 => sum f()(2,x) = 1

By Integral definition and triangle definition sum f()(2,x) = (x-2)^2/2
So (x-2)^2/2 = 1
... (x-2)^2 = 2
... x-2 = sqrt(2)
Finally x = 2 + sqrt(2)

b)
Let F(x) = 2 - x
Suppose sum f() (0,2) - sum f() (2,x) = F(x)
So f() (0,2) - sum f() (2,x) = 2 - x
... 1 - sum f() (2,x) = 2- x
... -sum f() (2,x) = 1-x
... sum f() (2,x) = x - 1
By Integral definition and triangle definition sum f()(2,x) = (x-2)^2/2
So sum (x-2)^2/2 = x - 1
... (x-2)^2 = 2x - 2
... x^2 - 4x + 4 = 2x - 2
... x^2 - 6x + 6 = 0
Finally by quadratic formula x = 3 - sqrt(3) and 3 + sqrt(3)

Nice bait user, freshman year stuff.

That was his homework you stupid fag.
Now leave and never return.

>The prof said "You just hit a nail with a sledgehammer" meaning it was technically accurate but the aspect I needed was a fairly trivial two line independent proof.
I love hitting nails with sledgehammers though.

Well, isn't integral just an area under graph?
A.
So, F(0) = 0, F(1) = 0.5, F(2) = 1, F(3) = 0.5
F(x) = 0 for x = 0 and some 3 < x < 4
(3, -1), (3, 0), (x, 0), (x, 2 - x) creates right trapezoid with of area 0.5
0.5 = (1 + x-2) / 2 * (x - 3)
(x-3)*(x-1) - 1 = 0
x^2 - 4x + 2 = 0
x = 2 + sqrt(2)

B.
looking for x where F(x) == f(x)
x = 0, obviously.
(1, 0), (1, 1), (x, 2 - x), (x, 0) creates right trapezoid with of area - |x - 1| + 1 - 0.5
Because x > 1, area = - (x - 1) + 1 - 0.5 = -x + 1.5
-x + 1.5 = (1 + 2 - x) / 2 * (x - 1)
0 = 1/2 (3-x) (x-1)+x-3/2
0 = -x^2/2+3 x-3
x = 3-sqrt(3)
x = 3+sqrt(3)

CS student here. Is that correct?

>And then there is Annabelle McAllister who is raising up the dead...

OP here, no it wasn't. You shouldn't be getting stuff like this as homework if you aren't underage.

People who do Calc I are usually 18+.

yeah! i managed to solve this, although had nothing to do with maths for a very long time and was rather mediocre at it. i spent like 45 minutes and wasted approximately 20 a4 sheets. (without calculator or looking it up of course)

(solution for b is also 0)

Phew. Feeling a bit rusty, but I think I might've done it.
a): x=0 or x=2+sqrt(2)
b): x=0 or x=3+-sqrt(3)
(Work in pic related)

Also, remind me to actually learn LaTeX before I do something like this again. Writing the equations directly into the post looks better and is more practical than taking screenshots of fucking OpenOffice Math /selffacepalm.