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relative extrema
- Thread starter newuser
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D
Deleted member 4993
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for relative extrema y' = 0 → y' = 2x*(x-a) + (x2-4) = 0 ...............now what.............y = (x + a)(x^2-4), a€R, has maximum and minimum relatives and maximum relative reaches at x=-1/3 . find the abscissa of minim relative.
Pls help me to solve this question. THANKS
find the roots of x.for relative extrema y' = 0 → y' = 2x*(x-a) + (x2-4) = 0 ...............now what.............
2x*(x+a) + (x2-4) = 0
3x^3+2ax-4=0
as it is quadratic in x it has 2 roots.
one roots is given that is x=-1/3.
now how to find other root?
D
Deleted member 4993
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find the roots of x.
2x*(x+a) + (x2-4) = 0
3x^3+2ax-4=0
as it is quadratic in x it has 2 roots.
one roots is given that is x=-1/3.
now how to find other root?
Find those roots as a function of 'a'...........................................(1)
Set one of those = - 1/3 and you can calculate the value of 'a'.
Now use (1) to find the other root.
A perhaps simpler approach than the one suggested by Subhotosh Khan is simply to insert - (1/3) into your quadratic and then to solve the resulting linear equation for a.find the roots of x.
2x*(x+a) + (x2-4) = 0
3x^3+2ax-4=0
as it is quadratic in x it has 2 roots.
one roots is given that is x=-1/3.
now how to find other root?
You can then replace a in the quadratic with a number and solve that quadratic to get the other solution for x.
There is no difference logically between the two approaches, but the algebra in my approach may seem simpler or more intuitive.
Please note: solving the quadratic does not answer the question posed.
Thanx for ur great help.A perhaps simpler approach than the one suggested by Subhotosh Khan is simply to insert - (1/3) into your quadratic and then to solve the resulting linear equation for a.
You can then replace a in the quadratic with a number and solve that quadratic to get the other solution for x.
There is no difference logically between the two approaches, but the algebra in my approach may seem simpler or more intuitive.
Please note: solving the quadratic does not answer the question posed.