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f(x)=x^3 + mx^2 +nx +p

solve m, n, p and f has to admit local extremes point in x=-1 and x=1

2007-04-18 04:14:19 · 3 answers · asked by mistikaaaaa 1 in Science & Mathematics Mathematics

3 answers

m=7
N=32
P=61.221
F=1.10102

2007-04-18 04:16:52 · answer #1 · answered by Anonymous · 0 0

for x = -1 and x = 1 to be extrema, need them to be solutions of f'(x) = 0, so

f'(x) = 3x² + 2mx + n = 0, and if x = -1,
3 - 2m + n = 0,
2m - n = 3

and if x = 1,
3 + 2m + n = 0
2m + n = -3

adding the 2 equations, 4m = 0, m = 0 (which makes the cubic an odd function with symmetry about the origin consistent with x = ±1 for extrema).

subtracting the 2 equations, 2n = -6, n = -3.

so we have f(x) = x^3 - 3x + p, and p can be anything we want, since it just translates the graph vertically, changing the y values of the extrema but not the x values.

2007-04-18 11:28:04 · answer #2 · answered by Philo 7 · 0 0

local extremes at x means that f'(x) = 0

f'(x) = 3 x^2 + 2mx + n
f'(-1) = 0 => 0 = 3 - 2m +n
f'(1) = 0 => 0 = 3 +2m + n
=> m = 0 and n = -3

It works for any p.

2007-04-18 11:21:57 · answer #3 · answered by roman_king1 4 · 0 0

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