Answer:
f(x) has a global minimum at x = -1/2
Step-by-step explanation:
Find and classify the global extrema of the following function:
f(x) = 2 x^2 + 2 x + 10
Find the critical points of f(x):
Compute the critical points of 2 x^2 + 2 x + 10
To find all critical points, first compute f'(x):
d/( dx)(2 x^2 + 2 x + 10) = 4 x + 2
= 2 (2 x + 1):
f'(x) = 2 (2 x + 1)
Solving 2 (2 x + 1) = 0 yields x = -1/2:
x = -1/2
f'(x) exists everywhere:
2 (2 x + 1) exists everywhere
The only critical point of 2 x^2 + 2 x + 10 is at x = -1/2:
x = -1/2
The domain of 2 x^2 + 2 x + 10 is R:
The endpoints of R are x = -∞ and ∞
Evaluate 2 x^2 + 2 x + 10 at x = -∞, -1/2 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
-1/2 | 19/2
∞ | ∞
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
-1/2 | 19/2 | 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
-1/2 | 19/2 | global min
f(x) = 2 x^2 + 2 x + 10 has one global minimum:
Answer: f(x) has a global minimum at x = -1/2
