we know that cos(x) intersects sin(x) in the interval 0< x< pi/2... but how can i find (x,y) points of this intersection? (without plotting and estimating based of the resultant graph)
it says give your answer "accurate to three decimal places" that is what I am unsure about... that sounds like i can just do one repetition and write the answer to 3 decimal places.. but that sounds too easy (its for my finals, it should be harder i think)
Homework Statement
using Newtons method with an initial estimate of x0=2, find the point where the graph f(x)=x3-x-2 crosses the x-axis
Homework Equations
xi+1 = xi - f(xi)/f'(xi)
The Attempt at a Solution
Using a function plotter, I know the answer should be around 1.52138...
we have z = 5+2i
how do i find the following:
|z|
z-1
i can do the basic operations (x, /, +, -) with complex numbers but i have no idea where to even start with these 2.
actually.. i just noticed something when i actually went back and wrote it... shouldn't it really be (x+(1-2i))(x+(1+2i))? because x = - 1 is x + 1 = 0 for both solutions so the only thing that should change in the outcomes is the sign in front of the 2i
why does one of the factors in your...
there is an example in my notes where the quadratic equation has been used for polynomials with no real solutions. Using the same principal here we have:
x = (-2 +- sqrt(-16)) / 2
now, the square root of 16 is 4 so we have
x = (-2 +- 4i)/2
= -1 +- 2i
and that's how i got my roots.
i jus started in complex numbers today and that's what i came up with.. if someone could please move the thread to the correct place that would be appreciated..
Mark44, Can you please advise how i should go about finding the solution to that part of the question?
x3+4x2+9x+10
finding 1 root and using synthetic division we can factorize to:
(x+2)(x2+2x+5)
using complex numbers to factorize (x2+2x+5) we have (x+1)(x-2i)(x+2i), and so our final solution is:
(x+2)(x+1)(x-2i)(x+2i)
is this correct?
i was aware of the definition i=sqrt(-1) but didnt notice it in there. thanks for pointing that out.
i am also in aus, i am doing undergrad degree in applied finance at macquarie, its for math130 which is roughly equivalent to 3 unit hsc math.
i have just started on complex numbers today and have read that the "algebraic rules for complex are the same ordinary rules for real numbers"..
when multiplying 2 complex numbers (z1 and z2) i can see easily that:
(x1+y1i)(x2+y2i) = x1x2 + y1x2i + x1y2i +y1iy2i
however I am struggling to...
ok, so using C = -1/2 we get v(0) = -1, and using C = 1/2 we get v(0) = 0
neither of which gets our true initial velocity of -1/2... the only way i found to get the correct initial velocity at t(0) is using C = 0.
thanks, i can see how that is true. However i can't put it all together.
The object is moving at v = -1/2 at time 0, from a starting point of 5 on a grid.. how can i find out where the object is 5 seconds later?
Is it the integral of the integral of acceleration (sin2t) where t = 5?
Im trying to find the position of an object x at time t. There is pretty straight forward formulas to use to find x(t), however acceleration must be fixed... but in this question acceleration is changing at sin2t.
I have been going through period by period working out the new position at...
0 = 108x2-66x+6
0 = 108x2-54x-12x+6
0 = 54x(2x-1)-6(2x-1)
6(2x-1) = 54x(2x-1)
6 = (54x(2x-1)) / 2x-1
6 = 54x
x = 1/9
sorry i meant 1/9 not 9
is this still incorrect? as u said .. there should be 2 of them..
Homework Statement
Using the local minima, local maxima and points of inflection of the following function, plot the graph:
f(x) = 9x4-11x3+3x2+1
The Attempt at a Solution
f(x) = 9x4-11x3+3x2+1
f ' (x) = 36x3-33x2+6x
= 3x(12x2-11x+2)
= 3x(3x-2)(4x-1)
therefore we have x...
Finding the "shortest ladder"
Problem
I was given an analogy involving a ladder that goes over a fence and then leans against a wall a meter after the fence. The question wanted me to answer "what is the shortest ladder that goes over the fence and reaches the wall"
The Attempt at a Solution...
Here are several problems (differentiation) i have attempted but not completely sure if they are correct
Homework Statement
differentiate y = sec(ex)
The Attempt at a Solution
y = sec(ex)
y = sec(u)
(where u = ex)
dy/dx = tan(u)sec(u)ex
tan(ex)sec(ex)ex
If this is correct so far...
Homework Statement
derivative of esec(x)
The Attempt at a Solution
u = sec(x)
y = eu
du/dx = tan(x)sec(x)
dy/du = eu
dy/dx = dy/du * du/dx
= esec(x)tan(x)sec(x)
can someone show how to simplify \frac{12x+9}{x} = x+9 so that it is in the quadratic form: x^2 -3x -9 = 0
multiplying both sides by x, subtracting both sides by 9 etc. . doesn't give me the quadratic.
for g(x) = (x-2) / (x^2 -9)
the domain of the function would be simply to have the denominator not equal to 0, which in this case would be x = 3.
but.. the solution states that it is -3 and 3 cannot be used.
this is confusing seeing that -3^2 -9 = -18 which is not 0.
maybe it...
can i also do this for f(x) = (x^2 -1) / x ?
(for this question it doesn't say which rule to use)
or should i just use the quotient rule for this?
i mean, i should get the same answer if i use the quotient rule or the definition of a derivative for this one?
maybe break it up into f (x) = x +2 and g (x) = x - 2, find the derivatives of f(x) and g(x) using the definition, then solve f '(x) / g '(x)?
is that what the question means?
find all solutions to sin^2 x = 1/4 in pi/2 <= x < 2pi (i.e in quadrants 2,3 and 4)
i understand what i need to do but don't understand sin^2 part. will i need to apply pythagoras therom here first?
thanks
Im trying to find the derivative of f(x) = x+2 / x-2
I know the formula to apply to this but it get quite messy because this example is a fraction.
Maybe i need to put function f(x) in a more simplier form before attempting to find its derivative?
Just trying to find a way to work out the trig ratios for angles with large fracetions in the unit circle (e.g. sin(15pi/2) etc..)
for angles with smaller fractions like cos(-7pi/4) i can solve easily like this: 7/4 = 1.75 = 45 degree (pi/4) angle in the 1st quadrant (because its negative)...
isnt that obvious? u should trust the student to do the responsible thing and look for the concepts in the solution first. Especially when the question is of such horrid simplicity.
instead all i got was more questions as if its an interactive-learning-special-ed-class.
and there's...
Thread after thread after thread I am getting the same thoughtless comments.
I do take math seriously. My degree is based on math. I am a 20 year old second year Applied Finance / App Statistics student at top 8 Australian uni, apart form this algebra am currently undertaking I am also...
u might of misunderstood what i meant.. what i meant was: "just post the solution (as i will understand it without additional commentary) and so i can get on with applying it to similar examples " :)
thanks kenewbie, that gives me something to experiment with. Also, I find that playing around with these rules and just experimenting with concepts in general is making Algebra more and more natural to me ;)