Find the minimum or maximum value of the function.
h(x) = 3x2 – 18x+ 25

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:

 f(x) has a global minimum at x = 3

so what is the problem what are they asking you to solve

step-by-step explanation:

so what is the problem what are they asking you to solve

explanation:

step-by-step explanation: