Finding max. and min. values analytically

[lp27]

Homework Statement

y = 2x^2-14x

Homework Equations

The Attempt at a Solution

y' = 4x - 14
4x - 14 = 0
x = 14/4
f(14/4) = 2(14/4)^2 - 14 (14/4) = -49/2

I don't know how to continue from there. ((14/4), (-49/2)) is my critical point?

 

[wisvuze]

Homework Statement

y = 2x^2-14x

Homework Equations

The Attempt at a Solution

y' = 4x - 14
4x - 14 = 0
x = 14/4
f(14/4) = 2(14/4)^2 - 14 (14/4) = -49/2

I don't know how to continue from there. ((14/4), (-49/2)) is my critical point?

there are a couple of steps involved with finding local min/max's:

1) differentiate
2) find f'(x) = 0
3) take values larger and smaller than x | f'(x) = 0 to see if it is a min/max [note: you must do this because you may have f'(x) = 0 without being a min/max]
min: \ /
max : / \
meaning, values less than x | f'(x) = 0 will have negative derivatives if it is a minimum value.. et c .

 

[Mark44]

You can also use the second derivative to determine a local maximum or local minimum. At any critical point c, if f''(c) > 0, there is a local minimum at (c, f(c)). If f''(c) < 0, there is a local maximum at (c, f(c)).

For your particular function, you could also notice that its graph is a parabola that opens upward. The only critical point is at the vertex of the parabola.

BTW you should simplify 14/4 to 7/2.