# user667648

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bio website location Berkeley, CA age member for 3 years seen 12 hours ago profile views 121

 Jun29 revised Algebra and quadratic equations Added latex. Jun29 comment Algebra and quadratic equations What do you think the answer is? Jun29 suggested suggested edit on Algebra and quadratic equations Jun28 comment Find $G'\left( x\right)$. You don't want G^', you just need G'... That will stop the latex error. Jun28 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. Jun28 comment When are there no critical points? @Gahawar: Wouldn't that function have a critical point everywhere? Since there exists no derivative? Jun28 revised Why is the function integrable? added 2 characters in body Jun28 revised Why is the function integrable? added 6 characters in body Jun28 answered Why is the function integrable? Jun27 comment What do I not understand about one-to-one functions? @JBKing: Sorry I meant -1... Typed it too quickly. Jun27 accepted What do I not understand about one-to-one functions? Jun27 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! Jun27 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? Jun27 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? Jun27 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. Jun27 revised What do I not understand about one-to-one functions? added 1 character in body Jun27 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? Jun27 asked What do I not understand about one-to-one functions? Jun12 suggested suggested edit on Calculus optimization quick question Jun11 answered How do I solve the following absolute value equation?