calc
posted by kooldude .
find the values of a nad b if the function f(x)=2x^3 + ax^2 + bx + 36 has a local max when x=4 and a min when x=5
First you calculate the derivative:
f'(x)=6x^2 + 2ax + b (1)
At the local maximum and minumum f' is zero. If a polynomial is zero at some point p, then it must contain a factor (xp). So, since you know that f' is zero at both x =4 and x = 5 you know that f' must be of the form:
f'(x) = A(x+4)(x5) (2)
the two factors (x+4) and (x5) make the right hand side a second degree polynomial, so A must be a constant. If f' were a third degree function then A would have been an unknown linea function. From (1) and (2) you find
6x^2 + 2ax + b = A (x+4)(x5)
The coefficient of x^2 on the left hand side is 6, on the right hand side it is A, so you find that A = 6.
6 (x+4)(x5) = 6 x^2 6 x 120,
so 2a = 6 > a = 3
and b = 120.
thanks!
it would help if you wrote the quistionbozo and by the way YOUR NOT KOOL
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