Find the maximum value and minimum value of the function.

Sep 2012
37
0
NYC
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),
 
Oct 2012
259
19
israel
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.
 
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Dec 2011
2,314
916
St. Augustine, FL.
...
Also, i am not sure what is meant by the closed interval (-4, 4)...
A closed interval, denoted with brackets, such as \(\displaystyle [a,b]\) means you include the end-points, i.e., \(\displaystyle a\le x\le b\).

An open interval, denoted with parentheses, such as \(\displaystyle (a,b)\) means you do not include the end-points, i.e., \(\displaystyle a<x<b\).
 
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Sep 2012
37
0
NYC
A closed interval, denoted with brackets, such as \(\displaystyle [a,b]\) means you include the end-points, i.e., \(\displaystyle a\le x\le b\).

An open interval, denoted with parentheses, such as \(\displaystyle (a,b)\) means you do not include the end-points, i.e., \(\displaystyle a<x<b\).
So do I have to do anything with the points (-4,4) like plug it somewhere into the equation or am I done?
 
Dec 2011
2,314
916
St. Augustine, FL.
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.
 
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Sep 2012
37
0
NYC
How did I do this one?


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.