SOLUTION: let {{{ f(x) = ax^3 + 6x^2 + bx + 4 }}}. Determine the constants a and b so that f has a local minimum at x = -1 and a local maximum at x = 2. Show all your work using calculus.
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-> SOLUTION: let {{{ f(x) = ax^3 + 6x^2 + bx + 4 }}}. Determine the constants a and b so that f has a local minimum at x = -1 and a local maximum at x = 2. Show all your work using calculus.
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Question 1163320: let . Determine the constants a and b so that f has a local minimum at x = -1 and a local maximum at x = 2. Show all your work using calculus.
Thank you. Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52781) (Show Source):
Local minimum and local maximum x-values are the roots of the derivative
f'(x) = 0 = 3ax^2 + 12x + b (1)
So, you have this system of two equations for two unknowns "a" and "b"
f'(-1) = 3a*(-1)^2 + 12*(-1) + b = 0 (2)
f'(2) = 3a*2^2 + 12*2 + b = 0 (3)
or
3a + b = 12 (4)
12a + b = -24. (5)
Subtract equation (4) from equation (5) to get
9a = - 36, a = -36/9 = -4.
Then from equation (4),
3*(-4) + b = 12, b = 12 + 12 = 24.
ANSWER. a = -4, b = 24.
Visual check
Plot y = .
Done.
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I do not understand, for what need/reason tutor @greenestamps re-told my solution in his post.
It did not become more clear after that.
In any case, thanks for popularizing my solution (!)
A local maximum or minimum means the slope of the graph is 0; that means the derivative of the function is 0.
The given function is a 3rd degree polynomial; its derivative will be a 2nd degree polynomial. That is consistent with the given information that the function has one local minimum and one local maximum.
Find the derivative of the function:
Use the fact that the derivative is 0 at x=-1 and x=2 to form two equations in the unknowns a and b.
(1)
(2)
Subtracting (1) from (2)...
Then substituting a=-4 in either (1) or (2) gives b=24.
Do you feel slighted because I felt a solution identical to yours but presented differently might be useful to the student?!
Your solutions are NOT always the solutions that are the best possible for the particular student. The same solution presented differently might satisfy the student's needs better....
