quadratic equations

gourish · 2014-03-15 06:34:14

if x is real If x is real, then the minimum value of the minimum value of x^2+2x-2 over x^2-3x+3 is.... i am very poor at finding answers to such question can anybody help me and also suggest a way to answer such questions

Bob · 2014-03-15 08:25:50

hi gourish,

In view of your other post I'll do some graph sketching first:

I'd start by finding out how the quadratics graphs behave on their own.

(i) x^2 - 3x + 3. If you compute (-3)^2 - 4 x 1 x 3 [b squared minus 4ac] you'll see it has no real roots. So it never crosses the x axis and as it goes through (0,3) that means it is never negative. Differentiating shows a minimum at x = 3/2.

(ii) x^2 + 2x - 2. This has two roots, one negative, one positive. It has a minimum at x = -1. It goes through (0,-2).

(iii) Now for the function made up by dividing (i) by (ii). As x tends to infinity, this tends to x^2/x^2 = 1. The numerator will exceed the denominator so it drops towards 1 from above. As x tends to - infinity again it tends to 1, but this time from below.

(iv) The denominator is never zero so it does not have vertical asymptotes.

(v) When x = 0, the function is -2/3

This is enough to sketch the graph. Draw a line at y = 1. From - ∞ it drops down from this line, crossing the x axis at the negative root found in (ii). It re-crosses the x axis at the positive root found in (ii), rising to a maximum before dropping back towards y = 1.

So the minimum is somewhere between the two roots found in (ii).

Differentiate if you have to find the exact point.

This will be zero when the numerator is zero:

I've already identified the first of these as the minimum.

Bob

gourish · 2014-03-15 16:44:05

hi bob,
i really liked your suggestion i hope i will get better at finding answers to such questions i am curious
1.) if i can find the maxima and minima of such functions
2.)how can differentiating such functions help me plot graphs? because in the end part of reply you differentiated the function and after simplification found the answer as x=0 or x=2
3.)i tried to apply limit as x tends to 0 the function reaches -2/3 which luckily was the right answer as per my answer sheet. am i right? or is my answer sheet and me wrong?

Bob · 2014-03-15 19:54:14

hi gourish,

Differentiation will help you to find maxima and minima. Sometimes you can find them by other methods. Once you know these you can put on these as key points on the graph.

From my sketch, I already had a good idea that the minimum was close to x=0. Doing dy/dx told me it was exactly at x=0. I had already worked out that it would be a minimum and that y = -2/3 at that point.

For this function you don't need to let x 'tend to' 0. You can just substitute x=0 and get the value straight off.

Bob

gourish · 2014-03-16 03:09:03

well can you tell me how to find such maxima and minima's

P.S. if i differentiate "x^2+y^2=r^2" can i get "x*diff(x)+y*diff(y)=r*diff(r)" i haven't come across such differentiation yet. But saw one such differentiation but i am not sure if it's right

Bob · 2014-03-16 04:44:45

hi gourish,

When you differentiate something it has to be 'with respect to' something.

eg. y = 2x + 3

You differentiate with respect to x so

dy/dx = 2

In your example, you have a function of x AND a function of y, AND I expect r squared is a constant. So each part has to be done in a slightly different way.

You bring the power to the front and reduce the power by 1.

This would be 2y if we differentiated with respect to y. Because we are differentiating with respect to x you have to multiply by dy/dx

finally

Because this is not a variable, it differentiates to zero.

So, putting the whole thing together:

You can learn more at:

http://www.mathsisfun.com/calculus/deri … ction.html

Bob

gourish · 2014-03-16 16:03:38

hi bob
well is dependent on x and y well in fact this was an equation regarding the motion of an object in x-y plane in a physics problem i understood the rest of the part but somehow "x^2+y^2=r^2=x*diff(x)+y*diff(y)=r*diff(r)" this was the problem that was used to find the write answer in fact it was a question asked in JEE advanced of 2013 so i am curious that how is this differentiation possible. i assume that it's differentiation of each term with respect to time only one term at a time x^2 as x*x and diff(x*x)=x*diff(x) and of course diff with respect to time each time... is my thinking right and is it then mathematical correct.?

Bob · 2014-03-16 20:40:21

hi gourish

So it was for motion in a circle and with respect to time. You want d(x^2)/dt so you use the chain rule: = d(x^2)/dx times dx dt

which can be re-arranged to

The motion is at right angles to the radius line.

Bob

Math Is Fun Forum

#1 2014-03-15 06:34:14

quadratic equations

#2 2014-03-15 08:25:50

Re: quadratic equations

#3 2014-03-15 16:44:05

Re: quadratic equations

#4 2014-03-15 19:54:14

Re: quadratic equations

#5 2014-03-16 03:09:03

Re: quadratic equations

#6 2014-03-16 04:44:45

Re: quadratic equations

#7 2014-03-16 16:03:38

Re: quadratic equations

#8 2014-03-16 20:40:21

Re: quadratic equations

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