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Math Problem
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Denmark resolut 
1. Find interval on wich f is inc. and f is decr. f(x) = 12 arctan (5x) - 6 ln (25x^2+1) 2. for each x>0, a triangle is formed with vertices (0,0), (x,0), and (x,f(x)) where f(x) is the func. givn below. q: value of x which results in triangle of lrgest area? f(x)= 200 / (x^2 + 4x +25) PLS HELP MENS ))
2019-10-16 23:34
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#1
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Denmark resolut 
bump
2019-10-16 23:34
#2
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Germany Easyiwnl 
2+2=4-1=3
2019-10-16 23:35
2+2=/=4-1
2019-10-16 23:37
Quick maths
2019-10-17 00:29
f(x) = a^2 f(5) =5^2 f(5) = 25
2019-10-16 23:35
#4
ropz | 
Netherlands Sky3rr 
If those are normal Danish school problems, I understad why Xyp9x is this smart
2019-10-16 23:36
#5
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Germany Easyiwnl 
He is a robot
2019-10-16 23:37
lul
2019-10-16 23:37
#6
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Netherlands Removed 
those are normal Dutch school problems too u pannenkoek
2019-10-16 23:37
#11
ropz | 
Netherlands Sky3rr 
They are? Fuck I really should start studying
2019-10-16 23:38
Use graphing calculator desmos.com/calculator
2019-10-16 23:37
#13
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Denmark resolut 
well i have to show my work as this isn't a multiple choice question
2019-10-16 23:39
#16
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Netherlands Removed 
Seeing how the question is "find interval" and not "show algebraically" or "calculate precisely/exactly" it means you are allowed to do it with a graphing calculator and it doesn't have to be exact.
2019-10-16 23:43
Well you could graph rational function 200x/(x^{2}+4x+25) by hand for #2
2019-10-16 23:44
2145
2019-10-16 23:38
Do you know how to derivate?
2019-10-16 23:39
#14
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Denmark resolut 
yes. i'm asking because i'm not entirely confident with the answers
2019-10-16 23:41
I'm playing cs atm but if i feel like i may do it before sleeping. Can't promise anything.
2019-10-16 23:42
derivative-calculator.net/ This shows the intermediate steps so you can check where (if) you went wrong.
2019-10-17 00:00
f(x) = 12 arctan (5x) - 6 ln (25x^2+1) f'(x) = 60/(25x^2 +1) - 300x/(25x^2 +1) We set f'(x) = 0 to find points with 0 gradient, i.e. maximum points/minimum points/saddlepoints. 0 = 60/(25x^2 +1) - 300x/(25x^2 +1) => x= 1/5 f''(x) = 300(25x^2 -10x -1) / ((25x^2 +1)^2) Evaluating f''(1/5) = -150. Negative value tells us that at f(1/5) is at a maximum, from which we can deduce f(x<1/5) is increasing as x increases and f(x>1/5) is decreasing as x increases. I won't show the specifics of the calculation as a lot of this is useful for you to go through, especially for when you take exams. It involves the quotient rule and the product rule of differentiation. It also requires some knowledge of trig identities, namely tan(x) = sin(x)/cos(x), the derivative of sin(x) = cos(x), the derivative of cos(x) = -sin(x), the deritavtive of -sin(x) = -cos(x) and the derivative of -cos(x) = sin(x). It also requires knowing the derivative of ln(x), which is 1/x.
2019-10-17 00:00
You really use f''(x) to study crec and decrease in uk?
2019-10-17 00:09
Yes, how is it done in Spain?
2019-10-17 00:20
The second question is a little more interesting, and easier I would say. Firstly, we know the area of a triangle is given by 1/2*h*b where h=height and b=base. So for the triangle with vertices (0,0), (x,0), (x,f(x)), our triangle has area 1/2*x*f(x). Substituting in the given function f(x) yields: Area = a(x) = 100x / (x^2 + 4x +25). Differentiate with the quotient rule to yield: a'(x) = - 100(x^2 -25) / (x^2 +4x +25)^2 Set a'(x) = 0 to find points where gradient = 0, i.e. the maximum point we desire. a'(x) = 0 => x = +5 or -5. Already we can see x = -5 will not be our solution as the question states x>0, however we will check +5 to make sure it is a maximum, that is we will differentiate a'(x) and evaluate at a''(5), with the hope it yields a negative value. a''(x) = 200(x^3-75x-100) / (x^2 +4x +25)^3 a''(5) = -10/49. Negative value => maximum, as we desired, and expected. Finally evaluate at a(5) = 50/7. Conclusion: Largest area triangle of area 50/7 is achieved when x=5.
2019-10-17 00:18
It’s not more interesting. Both are boring and useless.
2019-10-17 00:26
I suppose you were solving differential equations already in your mother's womb.
2019-10-17 00:31
They talk about my differential equations
2019-10-17 00:33
#38
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United Kingdom Megamo10 
you have math degree mens
2019-10-17 00:59
xddddd high school use PhotoMath, its a app on android and IOS, its sick
2019-10-17 00:11
thats not a math problem thats a kindergarten problem
2019-10-17 00:11
Just differentiate, equal the result to 0 and solve for x
2019-10-17 00:23
a single "inequality of means" problem and thats just middle school math class. first, x^2+4x+25 = (x+2)^2+21 > 0, so f(x)= 200 / (x^2 + 4x +25) > 0. and, x + 25 / x >= 2(25x / x)^0.5 = 10, it gets only when x = 25 / x, or x = +-5 when x > 0. the area of the triangle S = xf(x)/2 = 100x / (x^2 + 4x +25) = 100/(x + 25/x +4) <= 100/(10+4) = 50/7, who gets equal sign when x=5 when x < 0. the area of the triangle S = -xf(x)/2 = -100x / (x^2 + 4x +25) = 100/[(-x) + 25/ (-x) - 4] <= 100/(10-4) = 50/3, who gets equal sign when x=-5 and since 50/7 < 50/3, we conclude as: x = -5 would results in triangle of largest area.
2019-10-17 00:33
#32
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Netherlands Removed 
>"middle school math" >2. for each x>0 >x = -5
2019-10-17 00:36
na education
2019-10-17 00:45
i dont like to read questions cuz we are testing math not english right?
2019-10-17 00:45
well for 2nd one is ez, triangle area will be x*f(x)/2 let's solve x*f(x)/2 -> max, s.t. x> 0 so basically d\dy((x *f(x))/2) = 0, x = +-5, so 5*f(5)/2 = 50/7 -5*f(-5)/2 = -50/3, x>0
2019-10-17 00:38
for the 1st prob: derivative (i guess it is obvious): f'(x) = (60-300x)/ (25x^2 + 1) , so denominator > 0 so we don't care bout it so 60 - 300x = 0 x = 1/5 f'(x) > 0 for all x < 1/5, so func is increasing, and f'(x) < 0 for all x > 1/5 so func is decreasing
2019-10-17 00:41
#36
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Finland Shall157 
I don’t really understand this stuff
2019-10-17 00:48
I’m so glad that I gave up on maths studies
2019-10-17 00:48
#39
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United Kingdom Megamo10 
link pic of question you tell it very badly
2019-10-17 01:01
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