373,754 views
0 votes
0 votes
NEED HELP ASAP pls and thx pic is attached

NEED HELP ASAP pls and thx pic is attached-example-1
User Rafael Aguilar
by
2.8k points

1 Answer

17 votes
17 votes

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)

NEED HELP ASAP pls and thx pic is attached-example-1
User Bogdan Evsenev
by
3.3k points