Mathematics: Post your doubts here!

whitecorp

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InnocentAngel

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whitecorp said:
Try squaring both sides and shift all terms to one side of the inequality before trying to work things out.

Hope this might help a little. Peace.
Click to expand...

I tried it. i got a quadratic inequality. Then from here i dont know what to do .
Please elaborate .
Thank you
 
N.M

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whitecorp said:
You would have to differentiate both sides wrt x,
ie dy/dx = 4x + (-4x^3)* (e^-x^4) and set dy/dx=0.
To obtain the roots of this equation, you will have to use your graphic calculator.
Subsequently, you can either use the second order derivative or the sign test method to ascertain the nature of the stationary values.

Hope this helps. Peace.
Click to expand...

but how to find the roots? this is where i was stuck... :-(
can't simplify it further after differentiating it...
my calculator doesn't have that option, n i think to show the working is imp instead of directly writing the roots from the calculator...
 
Taiyaba

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Show that the points (7,12), (-3,-12) and (14,-5) lie on a circle with centre (2,0).
Plz help
Thnx in advanc
 
whitecorp

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N.M said:
but how to find the roots? this is where i was stuck... :-(
can't simplify it further after differentiating it...
my calculator doesn't have that option, n i think to show the working is imp instead of directly writing the roots from the calculator...
Click to expand...

I guess you are not using a graphic calculator then?

If that is the case I am afraid you might have to consider other solving methods such as linear interpolation, newton raphson etc.

Or you could try approximating e^(-x^4) as a Maclaurin's Series (perhaps up to and including the term in x^4) to make things easier.

Hope this helps. Peace.
 
whitecorp

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Taiyaba said:
Show that the points (7,12), (-3,-12) and (14,-5) lie on a circle with centre (2,0).
Plz help
Thnx in advanc
Click to expand...

The equation of this particular circle is simply (x-2)^2 + y^2 =r^2, where r denotes the radius of the circle.
Substitute one set of coordinates into the above equation to compute the value of r, and subsequently substitute
the remaining two sets of coordinates to verify they satisfy the circle equation.

Hope this helps. Peace.
 
Taiyaba

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whitecorp said:
The equation of this particular circle is simply (x-2)^2 + y^2 =r^2, where r denotes the radius of the circle.
Substitute one set of coordinates into the above equation to compute the value of r, and subsequently substitute
the remaining two sets of coordinates to verify they satisfy the circle equation.

Hope this helps. Peace.
Click to expand...
(x-2)^2 + y^2 =r^2 is this a formula used to find the solution for these type of questions?
And it's still not v clear so can u please solve it?
Thnx
 
whitecorp

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Taiyaba said:
(x-2)^2 + y^2 =r^2 is this a formula used to find the solution for these type of questions?
And it's still not v clear so can u please solve it?
Thnx
Click to expand...

If a circle of radius r is centered at (a,b), then its cartesian equation is given by (x-a)^2+ (y-b)^2 =r^2 . In your question, a=2 while b=0.
Hope this helps. Peace.
 
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