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I have this problem from my Calculus 1 homework and I don't know if I'm doing it correctly.
f (x) = x^3 - 12x
For stationary points: f ' (x) = 0
f ' (x) = 3x^2 - 12 (solve for 0)
3x^2 = 12
x^2= 4
x= +2 and -2
(2, -16) and (-2, 16)
f '' (x) = 6x
f '' (2)= 6(2)
= 12
f '' (-2)= 6(-2)
= -12
Also, i am not sure what is meant by the closed interval (-4, 4),
http://i45.tinypic.com/258w0ba.jpg
evaluate the function at the two endpoints and at the stationary points inside the open interval. the largest(smalest) is the absolute maximum(minimum) of the function in the closed interval.
A closed interval, denoted with brackets, such asQuote:
Originally Posted by calculus123
means you include the end-points, i.e.,
.
An open interval, denoted with parentheses, such asmeans you do not include the end-points, i.e.,
.
So do I have to do anything with the points (-4,4) like plug it somewhere into the equation or am I done?Quote:
Originally Posted by MarkFL2A closed interval, denoted with brackets, such as
An open interval, denoted with parentheses, such as
Yes, you need to evaluate the function at the end-points to see if either are greater or smaller than the function's values at the critical points in the interval.
From these 4 points, choose the smallest as your absolute minimum, and the largest as your absolute maximum.
How did I do this one?
http://i48.tinypic.com/o6e1s1.jpg
f(x)= x³ + 4/x² + 7; x < 0
f'(x) = 3x² - 8x^-3
3x² - 8x^-3 = 0
x^5 = 8/3
x = (8/3)^(1/5) ≈ 1.21
Let's check what the slope is at x = -1
3 + 8 = 11
It is increasing over the entire interval.
f(-3): -27 + 4/9 + 7
The value at x = -3 is negative
and the value of f(-1) = -1 + 4 + 7 is positive.
Therefore, only one root.
bump.