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Can someone help me solve this problem:

Find values of a, b, c, and d such that g(x) = ax^3 + bx^2 + cx + d has a local maximum at (2, 4) and a local minimum at (0, 0). Thanks so much guys.... Luv ya :)

2007-03-25 00:09:43 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

@ local max. or min g(x)'=0
therefore
1st derivative of g(x)=3ax^2 + 2bx + c=0 @ x=2 & x=0

@ x=2;g(2)=4
g(x)= a(2)^3 + b(2)^2 + c(2) + d=4
= 8a+4b+2c+d=4-----(1)
since this is a min,g(x)'=0
g(2)'=3a(2)^2 + 2b(2) + c=0;
= 12a+4b+c=0-------(2)

@ x=0;
g(x)= a(0)^3 + b(0)^2 + c(0) + d=0-----(3)
g(x)'=3a(0)^2 + 2b(0) + c=0;-----(4)

from (3); d=0
from (4); c=0

(1)--->2a+b=1;
(2)---->3a+b=0;
So we have b=0,a=1


U better check the workings again, migth hav made a mistake ...but this is the method

2007-03-25 00:22:27 · answer #1 · answered by Tharu 3 · 0 0

g(x) = ax³ + bx² + cx + d, so differentiating yields

g'(x) = 3ax² + 2bx + c.

Since there are local turning points at x = 2, 0, we know that g'(0) = 0 and g'(2) = 0. Thus g' is a quadratic with those roots, so we can write it as

g'(x) = d(x - 2)(x - 0) = dx² -2dx,

for some constant d. Equating coefficients yields that

d = 3a, - 2d = 2b and c = 0,

so we can rewrite g as

g(x) = ax³ - 3ax² + d.

We know

g(0) = d = 0 and

g(2) = 8a - 12a = 4, ie. a = -1.

Hence,

g(x) = -x³ + 3x².

2007-03-25 07:42:19 · answer #2 · answered by MHW 5 · 0 0

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