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.
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),
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.
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?
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.