Calc. Limits

[Chaosniper23]

A) Verify that max {f(x), g(x)} = 1/2 {[f(x)+g(x)] + |f(x)-g(x)|}.

B) Find a similar expression for min {f(x), g(x)}

Not even sure how to approach this. Help will be greatly appreciated. Thanks a bunch.

 

[mmm4444bot]

To me, both parts of this exercise seem to be testing understanding of function notation and symbolic reasoning. Not much more.

There are different ways to prove stuff. I'll take a graphical approach.

The verification in (A) is obvious, if you draw a picture of two arbitrary functions (with one above the other at least part of the time) and reason out the given statement, piece-by-piece.

Plot two random curves in Quadrant I.

Label them f(x) and g(x), so you have something to look at, if you start to get confused about what curve is what.

Now, pick some arbitrary point on the x-axis, in an interval where the graph of f(x) is above the graph of g(x), and label this location c.

Draw the vertical line x = c such that it passes through both curves.

Put dots at each of the two intersection points.

On the curve of function f, label the dot f(c).

On the curve of function g, label the dot g(c).

Label, with the name b, the distance of the vertical line segment that runs from the x-axis to the point at g(c).

In other (mathematical) words: g(c) = b

Label the distance of the vertical line segment that runs from g(c) to f(c) with the name a.

In other "words": f(c) - g(c) = a

Now you have a diagram sufficient to perform the verification in part (A).

max{ f(x), g(x) } is obviously f(x) when x is c, yes?

After all, that's how I instructed you to draw and label the picture.

Therefore (at our example value for x), max{ f(x), g(x) } = a + b

The expression f(x) + g(x) equals a + 2b.

You follow that ? If not, look at your diagram.

The expression |f(x) - g(x)| is simply a.

Okay ?

Therefore:

max{ f(x), g(x) } = a + b

1/2 [f(x) + g(x) + |f(x) - g(x)|] = 1/2(2a + 2b) = a + b

Verified, for the case where f(x) > g(x).

To verify the case where f(x) < g(x), realize that f and g are just names and thus the case where g(x) > f(x) is already verified.

I mean, you could simply switch the function labels on your diagram, and the end result will clearly be the same.

You picked the location of c arbitrarily. The reasoning is that c could be any Real number within any interval in the domain of each respective function (where they both exist) according to the conditions described in the diagram.

Also, the actual values of f(c) and g(c) are not relevant. It's their relative positions that are relevant. Those locations could be in any Quadrant.

So, the diagram covers all Real x = c over the entire xy-plane, so we've covered all the bases.

The case where f(x) = g(x) is trivial because a = 0 and b = 0.

There are no other cases to consider, but, if you want to think about when either f(c) or g(c) is on the x-axis or the x-axis passes between them, just remember that a and b are distances (always positive).

If you've understood this far, then you should understand what they want for (B).

If you have not understood this far, then figure out where you first get confused and please post specific questions.

Cheers ~ Mark