Asked by Sandra
How do you find the critical numbers and the end points of this function:
f(x)=2x^3-3x^2-12x+5
i calculated the critical numbers as x=-1 and x=2 is this right?!?!
f(x)=2x^3-3x^2-12x+5
i calculated the critical numbers as x=-1 and x=2 is this right?!?!
Answers
Answered by
Reiny
great so far.
now find f(2) and f(-1) for the y values of your critical points.
unless there is a domain given for your x's , there are no "end points" for your function.
now find f(2) and f(-1) for the y values of your critical points.
unless there is a domain given for your x's , there are no "end points" for your function.
Answered by
drwls
If by "ritical numbers" you mean where the function has a local maximum or minimum, that would be where
f'(x) = 6x^2 -6x -12 = 0
x^2 -x -2 = 0
(x+1)(x-2) = 0
yes, -1 and 2 are correct.
f(x) has no end points or absolute maximum or minimum. It goes to + and - infinity when x does the same
f'(x) = 6x^2 -6x -12 = 0
x^2 -x -2 = 0
(x+1)(x-2) = 0
yes, -1 and 2 are correct.
f(x) has no end points or absolute maximum or minimum. It goes to + and - infinity when x does the same
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