Afunction defined on an interval [a, b] or (a, b) is uniformly continuous if for each ϵ> 0 there exists a δ> 0 such that |x−t|< δ implies that |f(x)−f(t)|< ϵ. then it gives a little note saying that δ cannot depend on x, it can only depend on ϵ.with ordinary continuity, the δ can depend on both x and ϵ. i'm just a little lost on why |x−t|< δ implies |f(x)−f(t)|< ϵ, and how δ can't depend on x but only ϵ.

Step-by-step explanation:

The secret for understanding the uniform continuity is to understand the continuity in a point. Recall, we say that a function f is continuous at a point x_0 if and only if or each ϵ>0 there exists a δ>0 such that |x_0−t|<δ implies that |f(x_0)−f(t)|<ϵ.

Consider this point of view. We are studying the continuity of a function f at x_0. The definition means that if we fix an ϵ>0, no matter how small, we can find δ>0 such that all the points t at distance δ from x_0 have the its images f(t) at a distance at most ϵ from f(x_0). In ‘‘natural’’ language. If we separate a little from the point x_0, the images of the function f do not separate too much from f(x_0).

Now, notice that we are asking this condition for a fixed value x_0. What happens if check for the continuity of f at another point x_1? Given the same ϵ, will we find the same value for δ, or will be smaller?

The idea with uniform continuity is that the same δ works for the same ϵ, independently of the point x. While continuity is a local property (at a point), uniform continuity is a global property (in all a set or interval).

Check the attached images. The function is f(x)=1/x. Notice that the red band is for all the points (in the image) at a distance less than ϵ from f(x_i), i=0,1 and the black band is for the points at distance less than δ from x_i, i=0,1. In both images the red band has the same length (more or less).

Notice that the black band around x_0 is bigger than the black band around x_1. So, when we get closer to zero, the values of δ needed to keep the values of f(t) closer to f(x) become smaller. Hence, δ depends of ϵ and x.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3