Answer:
f(x) has a global minimum at x = 4.08628
Explanation:
Find and classify the global extrema of the following function:
f(x) = x^4 - 4 x^3 - 9 x^2 + x - 16
Find the critical points of f(x):
Compute the critical points of x^4 - 4 x^3 - 9 x^2 + x - 16
To find all critical points, first compute f'(x):
d/(dx) (x^4 - 4 x^3 - 9 x^2 + x - 16) = 4 x^3 - 12 x^2 - 18 x + 1:
f'(x) = 4 x^3 - 12 x^2 - 18 x + 1
Solving 4 x^3 - 12 x^2 - 18 x + 1 = 0 yields x≈-1.13994 or x≈0.0536696 or x≈4.08628:
x = -1.13994, x = 0.0536696, x = 4.08628
f'(x) exists everywhere:
4 x^3 - 12 x^2 - 18 x + 1 exists everywhere
The critical points of x^4 - 4 x^3 - 9 x^2 + x - 16 occur at x = -1.13994, x = 0.0536696 and x = 4.08628:
x = -1.13994, x = 0.0536696, x = 4.08628
The domain of x^4 - 4 x^3 - 9 x^2 + x - 16 is R:
The endpoints of R are x = -∞ and ∞
Evaluate x^4 - 4 x^3 - 9 x^2 + x - 16 at x = -∞, -1.13994, 0.0536696, 4.08628 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
-1.13994 | -21.2213
0.0536696 | -15.9729
4.08628 | -156.306
∞ | ∞
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
-1.13994 | -21.2213 | neither
0.0536696 | -15.9729 | neither
4.08628 | -156.306 | global min
∞ | ∞ | global max
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
-1.13994 | -21.2213 | neither
0.0536696 | -15.9729 | neither
4.08628 | -156.306 | global min
f(x) = x^4 - 4 x^3 - 9 x^2 + x - 16 has one global minimum:
Answer: f(x) has a global minimum at x = 4.08628