Staff Versus Members! (Current record: 200)

[anoni]

1.
I refuse to go to the mathematical representation of a god.
KEEP YOUR DEFINITION OF NOTHING TO YOURSELF!

[Midnight Madness]

2

And the further away we get the better.

[MrRazot]

0
one interesting piece of flawed logic i once saw
is that since x/0 = infinity (x being anything)
therefore x = infinity*0
which proves that god exists though some intricate metaphor that anything (x) = god(infinity*0)

[anoni]

1.
Well ok x / 0 doesn't actually necessarily equal infinity.
If you look at the graph of 1 / x, where x is a variable, it creates a hyperbola, which never actually reaches zero, but when it does it reaches both infinity and negative infinity in two points.

So it gets a bit trippy, but some people just call it infinity to get rid of the trippyness

Here's a question
f(x) = (x^2 - 9) / (x - 3)

What happens to f(x) as x approaches 3?

[MrRazot]

it doesn't
rather it just reaches one heck of a decimal
or am i wrong?
it would be rather sad to get it wrong as this is exactly what i'm studying right now for the exam next week
...
oh and 0

[anoni]

1
You're wrong actually, it approaches 6, NOW HOW DO YOU FIGURE THIS OUT

f(x) = (x^2 - 9) / (x - 3)
Which expands to
f(x) = (x + 3)(x - 3) / (x - 3)
cross out
f(x) = (x + 3)
as x approaches 3 f(x) approaches 6.

of course, when f(x) actually is 3... that's a different story

[MrRazot]

well then... 0... lol

[anoni]

1.
If you get a question like that in the exam, just declare 2.9999999999999999 as x in your calculator and it'll give you your answer xD

NEXT QUESTION

Solve for x

|x + 3| > 5

[MrRazot]

0


x>2
simple

[anoni]

1.
Half marks

These symbols "||" mean absolute value symbols. An absolute value symbol means that whether the answer is positive or negative doesn't matter, it will always be positive.

For example
|-3| = 3
|1 - 6| = 5

So there's actually two answers to
|x + 3| > 5

Answer 1:
x + 3 > 5
x > 2

Answer 2:
(You must make everything negative)
-x - 3 > 5
-x > 8
x < -8

Therefore the correct answer is
x < -8, x > 5


NEXT QUESTION:
this assumes you do calculus if you don't, ignore this question

Find the derivative of x²e^x

[MrRazot]

0
easy
2x^2e^(x-1)
no?

[anoni]

1.
Lets see:

u = x²
u' = 2x
v = e^x
v' = e^x

vu' + uv'

2xe^x + x²e^x
Or
xe^x(2 + x)

Sorry

Ok next question!

Integrate: ⌡(6x) / (x²) dx



(BTW: Don't let this demotivate you, I'm intentionally using tricky questions that are supposed to throw you on the wrong path )

[MrRazot]

0
like a true fox


aaand... cant be bothered to do that one

[anoni]

1.
Another trickish question.

You actually use logs
Remember that
⌡(f'(x)) / (f(x)) dx = ln(f(x))

So ⌡(6x) / (x²) dx
Take out the 3 so it is
3⌡(2x) / (x²) dx

And it integrates to...

3ln(x²)

NEXT QUESTION

Find the perpendicular distance between the point (0, 5) and the line 2x - y + 8 = 0

[MrRazot]

0.sqrt2