Is it always true that if; f(x) = f '(x), then; -f (x) = - \int f '(x) ?
That is, the negative of a derivative has an integral which is simply the negative of the original function? (ignoring arbitiary constants)
"The Greats" - and more.
[Edit: I have likely posted this in the wrong forum - any Mods are welcome to move it to a more apt location, apologies]
I'm forming a reading list (Undergrad) for myself comprising of modern texts and "classics" by those such as Euclid and Euler.
Advice often...
Homework Statement
For the function f(x) = 1/4 (x=2)^2 - 4, I'm confused with regards to finding x & y intercepts.
Homework Equations
The Attempt at a Solution
You can work them out from the function in its current form, i think? For y-int, let x=0, which leaves (0,-3)...
I haven't done Maths for some time, and my Algebra needs some work.
Homework Statement
Two questions in particular;
Homework Statement
Two questions in particular;
a) Rearrange for x in terms of y and t, to its simplest form.
t(t-x) = 3y(3y-x) [3y-t =/= 0]
b)...
The question as is written :
Solve for x
5cosx-2= -0.72 ( 0 \leq x \leq 2\Pi )
(The answer given is "1.31, 4.97".)
So, i done what seemed natural.
5cosx-2=-0.72
5cosx = 1.28
cosx = 1.28/5
cosx = 0.256
Firstly, I'm not even sure what unit my answer is supposed to...
Thanks, Tedjin. So, i see that
f(f(x)) = f\left(\frac{1-x}{x}\right) = \frac{1-\frac{1-x}{x}}{\frac{1-x}{x}}
Multiplying Numerator & Denominator by x would leave \frac{x-(1-x)}{1-x} = \frac{x+x-1}{1-x} = \frac{2x-1}{1-x}
Which may not seem to have been the method i was heading towards...
Hello people. This is a problem from an A-Level Math book (though i am independantly learning, have given myself 2 months to cram the whole book down my throat :-S ), so, apparently, it is simple. I understand the principles behind functions, this is more of a problem with understanding complex...