Answer:
4
Explanation:
Find and classify the global extrema of the following function:
f(x) = -x^3 + 5 x^2 + x
Find the critical points of f(x):
Compute the critical points of -x^3 + 5 x^2 + x
To find all critical points, first compute f'(x):
d/(dx) (-x^3 + 5 x^2 + x) = -3 x^2 + 10 x + 1:
f'(x) = -3 x^2 + 10 x + 1
Solving -3 x^2 + 10 x + 1 = 0 yields x = 1/3 (5 - 2 sqrt(7)) or x = 1/3 (5 + 2 sqrt(7)):
x = 1/3 (5 - 2 sqrt(7)), x = 1/3 (5 + 2 sqrt(7))
f'(x) exists everywhere:
-3 x^2 + 10 x + 1 exists everywhere
The critical points of -x^3 + 5 x^2 + x occur at x = 1/3 (5 - 2 sqrt(7)) and x = 1/3 (5 + 2 sqrt(7)):
x = 1/3 (5 - 2 sqrt(7)), x = 1/3 (5 + 2 sqrt(7))
The domain of -x^3 + 5 x^2 + x is R:
The endpoints of R are x = -∞ and ∞
Evaluate -x^3 + 5 x^2 + x at x = -∞, 1/3 (5 - 2 sqrt(7)), 1/3 (5 + 2 sqrt(7)) and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
1/3 (5 - 2 sqrt(7)) | -0.0490425
1/3 (5 + 2 sqrt(7)) | 21.9009
∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:
The open endpoints of the domain are marked in gray
x | f(x) | extrema type
-∞ | ∞ | global max
1/3 (5 - 2 sqrt(7)) | -0.0490425 | neither
1/3 (5 + 2 sqrt(7)) | 21.9009 | neither
∞ | -∞ | global min
Remove the points x = -∞ and ∞ from the table
These cannot be global extrema, as the value of f(x) here is never achieved:
x | f(x) | extrema type
1/3 (5 - 2 sqrt(7)) | -0.0490425 | neither
1/3 (5 + 2 sqrt(7)) | 21.9009 | neither
f(x) = -x^3 + 5 x^2 + x has no global extrema:
Answer: -x^3 + 5 x^2 + x has no global extrema