car+tree=good
Greycel
★
- Joined
- Oct 25, 2022
- Posts
- 22
- Online
- 5h 3m
car+tree=good?dog+foid+bed
car+tree=good?dog+foid+bed
probably, other factors like (speed) needs to be added or multiplied.car+tree=good?
@trying to ascend what accounts for 80, but only wants 20
you're god damn right
Solution?
930Solution?
Correct
Send differentiation questions or limits4x²/(1 - sqrt(1 + 2x))² < 2x + 9.
Find the values of x that satisfy this inequality
Ok:Send differentiation questions or limits
Yeah we're about to get into integration now. I give up what's the answerAre you practicing it?
2LYeah we're about to get into integration now. I give up what's the answer
Would you mind telling me how you did it?
This is false. Take e.g. f(x) = sin(x). Then sin(x)/x tends to 1 as x tends to 0, yet sin(x^2-1)/(x-1) tends to sin(-1)/(-1) = sin(1) != 2 as x tends to 0.
True, it was actually x ----> 1This is false. Take e.g. f(x) = sin(x). Then sin(x)/x tends to 1 as x tends to 0, yet sin(x^2-1)/(x-1) tends to sin(-1)/(-1) = sin(1) != 2 as x tends to 0.
RuhTrue, it was actually x ----> 1
Yeah, I realized that after I had made the post. In that case 2L is correct of course.True, it was actually x ----> 1
Would you still like to know or can you see it now?Would you mind telling me how you did it?
I don't get itYeah, I realized that after I had made the post. In that case 2L is correct of course.
Would you still like to know or can you see it now?
Multiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.I don't get it
ThanksMultiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.
npThanks
Find the critical points of (x² - x)/(1 + 3x²).Any basic (very) integration or differentiation questions?
y prime= (3x-1)(x+1)/[(1+3x^2)^2]Find the critical points of (x² - x)/(1 + 3x²).
Gave up but a=1/4 by l'hospital's ruleBe a = limx→0 of (13x - sin(10x))/(sin(11x) + x).
and b = limx→∞ x/sqrt(x² + x + 1).
Find a + b
5/4?Find a + b
Correct!y prime= (3x-1)(x+1)/[(1+3x^2)^2]
f(1/3)= -1/6
f(-1)=1/2
Gave up but a=1/4 by l'hospital's rule
Correct5/4?
Just used the chain rule only and got the wrong answerBe f(x) = sqrt(sin(7x + ln(5x))).
Find the first derivative of f(x)
3/4?Evaluate the limit
lim [x^2(3+sinx)]/[(x+sinx)^2]
x--->0
Nice but the question was introduced to me when I was learning how to use sinx/x=1 as x tends to 0 in questions3/4?
I just expanded the square in the denominator and then divided everyone by x²
334What's the biggest value of n, such that 21 (to the power of n), divides 2020! ?
3pi + 2Given that the integral (from 0 to pi) of (f(x) + sin(2x)) = 2.
Find the integral (from 0 to pi) of (f(x) + 9(x/pi)²)
Cos(7x + ln(5x)). (7 + 1/x) /2.sqrt(sin(7x + ln(5x)))Be f(x) = sqrt(sin(7x + ln(5x))).
Find the first derivative of f(x)
Using first principle, prove that cos'(x) = - sin(x).
First, plugging in y = 0 yields that f(x)*f(0) = f(x) which means that either f(x) = 0 for all x or f(0) = 1. One can easily check that f = 0 is not a valid solution however.Find all functions, R ---> R, such that:
f(x)f(y) = f(x + y) + xy.
e^2xSimplify the expression: I = (11/8) . e (to the power of 2x) + (3/4) . x² . e (to the power of 2x) - (3/4) . x . e (to the power of 2x) - (3/2) . integral of (x² . e to the power of 2x).