Answer:
f(x) has a global minimum at x = 3 and no global maxima
Step-by-step explanation:
Find and classify the global extrema of the following function:
f(x) = 3 x^2 - 18 x + 25
Find the critical points of f(x):
Compute the critical points of 3 x^2 - 18 x + 25
To find all critical points, first compute f'(x):
d/( dx)(3 x^2 - 18 x + 25) = 6 x - 18
= 6 (x - 3):
f'(x) = 6 (x - 3)
Solving 6 (x - 3) = 0 yields x = 3:
x = 3
f'(x) exists everywhere:
6 (x - 3) exists everywhere
The only critical point of 3 x^2 - 18 x + 25 is at x = 3:
x = 3
The domain of 3 x^2 - 18 x + 25 is R:
The endpoints of R are x = -∞ and ∞
Evaluate 3 x^2 - 18 x + 25 at x = -∞, 3 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
3 | -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
3 | -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
3 | -2 | global min
f(x) = 3 x^2 - 18 x + 25 has one global minimum:
Answer: f(x) has a global minimum at x = 3
