Answer:
It will open up.
Explanation:
It will open up.
The function has no maxima but minima at x=2.
Find and classify the global extrema of the following function:
f(x) = x^2 - 4 x + 2
Find the critical points of f(x):
Compute the critical points of x^2 - 4 x + 2
To find all critical points, first compute f'(x):
d/(dx) (x^2 - 4 x + 2) = 2 x - 4
= 2 (x - 2):
f'(x) = 2 (x - 2)
Solving 2 (x - 2) = 0 yields x = 2:
x = 2
f'(x) exists everywhere:
2 (x - 2) exists everywhere
The only critical point of x^2 - 4 x + 2 is at x = 2:
x = 2
The domain of x^2 - 4 x + 2 is R:
The endpoints of R are x = -∞ and ∞
Evaluate x^2 - 4 x + 2 at x = -∞, 2 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
2 | -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
2 | -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
2 | -2 | global min
f(x) = x^2 - 4 x + 2 has one global minimum:
Answer: f(x) has a global minimum at x = 2
Look at the graph; there is no x-intercept at (2,0 rather at x = 2 - sqrt(2) & x = 2 - sqrt(2))
The axis of symmetry is at (2,-2)
plane curve | Cartesian equation f(x) = x^2 - 4 x + 2 | axis of symmetry
line | through (2, -7/4) = (2, -1.75)
through (2, -2)