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visits member for 3 years
seen 12 hours ago

please delete me


Jun
29
revised Algebra and quadratic equations
Added latex.
Jun
29
comment Algebra and quadratic equations
What do you think the answer is?
Jun
29
suggested suggested edit on Algebra and quadratic equations
Jun
28
comment Find $G'\left( x\right)$.
You don't want G^', you just need G'... That will stop the latex error.
Jun
28
comment How to interpret integration of a discontinous function
I believe it has to do with (informally speaking): Rationals are countably infinite, irrationals are uncountably infinite... Hence, the point discontinuities of the rationals are insignificant in the end. Which is why the integral is zero. Someone correct me if I'm wrong.
Jun
28
comment When are there no critical points?
@Gahawar: Wouldn't that function have a critical point everywhere? Since there exists no derivative?
Jun
28
revised Why is the function integrable?
added 2 characters in body
Jun
28
revised Why is the function integrable?
added 6 characters in body
Jun
28
answered Why is the function integrable?
Jun
27
comment What do I not understand about one-to-one functions?
@JBKing: Sorry I meant -1... Typed it too quickly.
Jun
27
accepted What do I not understand about one-to-one functions?
Jun
27
comment What do I not understand about one-to-one functions?
@JBKing: And thus, two-to-two, would mean that given a point, for any other arbitrary point it must not produce the same output! And in the case of x^2 it does, give x = 1, y = 1, let x_2 = -1, y = 1 thus it is not injective!
Jun
27
comment What do I not understand about one-to-one functions?
OHH! I think I get it know, it's misleading because most people just think of a normal function for every f(x) there is a unique y, but it should actually be that plus doing a "horizontal line" test? So, while x^2 is a function, it is not one-to-one, because it fails the horizontal line test. Whereas x^3 is?
Jun
27
comment What do I not understand about one-to-one functions?
I'm still a little bit confused, so if I am understanding, "thought incorrectly ... f(x) can only have one value", $f(x_1)$ can give multiple values, but $f(x_2)$ must give different values than $f(x_1)$ (given $x_1 \neq x_2$)? But one-to-one, does not imply that given one $x$, $f(x)$ gives one output?
Jun
27
comment What do I not understand about one-to-one functions?
Whoa, so an injective function is not necessarily a function, in terms of function vs relation? That is misleading. My high school terminology deceives me.
Jun
27
revised What do I not understand about one-to-one functions?
added 1 character in body
Jun
27
comment What do I not understand about one-to-one functions?
@PeterFranek: How so? I am not really sure what you want cleaning up on?
Jun
27
asked What do I not understand about one-to-one functions?
Jun
12
suggested suggested edit on Calculus optimization quick question
Jun
11
answered How do I solve the following absolute value equation?