What is the vertex if g(x) = 8x2 - 64x?

Find and classify the global extrema of the following function:

f(x) = 8 x^2 - 64 x

Find the critical points of f(x):

Compute the critical points of 8 x^2 - 64 x

To find all critical points, first compute f'(x):

d/( dx)(8 x^2 - 64 x) = 16 x - 64

= 16 (x - 4):

f'(x) = 16 (x - 4)

Solving 16 (x - 4) = 0 yields x = 4:

x = 4

f'(x) exists everywhere:

16 (x - 4) exists everywhere

The only critical point of 8 x^2 - 64 x is at x = 4:

x = 4

The domain of 8 x^2 - 64 x is R:

The endpoints of R are x = -∞ and ∞

Evaluate 8 x^2 - 64 x at x = -∞, 4 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

4 | -128

∞ | ∞

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

4 | -128 | 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

4 | -128 | global min

f(x) = 8 x^2 - 64 x has one global minimum:

 f(x) has a global minimum at x = 4

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse