What does this mean?

Question

A) local min at x=4
B)critical point x=2
C) f ' (x) > 0 on (-1,2)
How do I know if it shows on the graph or how does it show on the graph? I cant post a graph up here but let me describe it. It has an closed circle at (4,5) open circle at (4,4). The V (narrow part) is at (2,2). Plz help if u can. Thnx!

for part that I need the most ...how does that look on a function of a graph? My graph has a 2 holes. One is a filled-in circle at (4, 5) . The 2nd one is a unfilled-in circle at (4, 4). Would my graph have this property ...local min at x=4?

Answer 1

A) Local min at x = 4

This means f'(4) = 0, and
f''(4) > 0

B) If x = 2 is a critical point, it means that f'(2) = 0 or f'(2) is undefined. (Critical points are values of x such that f'(x) = 0 or f'(x) is undefined).

If x is defined, then the tangent at that point is a horizontal line. It also means that at that point, there *could* be a minimum, maximum, or neither.

C) f'(x) > 0 on (-1, 2) means f(x) is increasing on (-1, 2)

Answer 2

I'm not sure what you want to know... a local min, means the lowest y value on a differentiable closed interval
[4,infinity)

b) critical point at x=2 most critical points are local mins and maxes when derivative equals 0, but only a critical point and not a min/max when the derivative is undefined, but when its a V shape, meaning sharp corners, it is not differentiable, so it is still a critical point, but it is a point where derivative is undefined.

C) f'(x)>0 just means that the derivative is positive from x=-1 to x=2. this just means your graph of f(x) has a positive slope over that interval.