Finding Max & Min of f & g: A Comparison

[jdz86]

Homework Statement

(a) Let f,g: [a,b] \rightarrow \Re.

Define: f \vee g(x) = max(f(x),g(x)), x\in [a,b]
f \wedge g(x) = min(f(x),g(x)), x\in [a,b]

(b) Let f_{+} = f\vee0, f_{-} = -(f\wedge0)
Show that: f = f_{+} - f_{-}
abs value of f = f_{+} + f_{-}

Homework Equations

f_{+}, f_{-} \geq 0

The Attempt at a Solution

(a) f \vee g(x) equals the supremum and infimum for f \wedge g(x). Supremum would be "b" for both f and g, and infimum of both would be "a"??

(b) Lost with this one. It relates to the first question I know, but trying to put them together hasn't been working.

 

[HallsofIvy]

Homework Statement

(a) Let f,g: [a,b] \rightarrow \Re.

Define: f \vee g(x) = max(f(x),g(x)), x\in [a,b]
f \wedge g(x) = min(f(x),g(x)), x\in [a,b]

(b) Let f_{+} = f\vee0, f_{-} = -(f\wedge0)
Show that: f = f_{+} - f_{-}
abs value of f = f_{+} + f_{-}

Homework Equations

f_{+}, f_{-} \geq 0

The Attempt at a Solution

(a) f \vee g(x) equals the supremum and infimum for f \wedge g(x). Supremum would be "b" for both f and g, and infimum of both would be "a"??

NO, of course not. a and b are the smallest and largest values of x. Your functions are defined as inf and sup of f(x) and g(x), the function values.
What exactly are you trying to do here? In (a) you are given two definitions but I see no question!

(b) Lost with this one. It relates to the first question I know, but trying to put them together hasn't been working.

Again, what was the first question? What is f₊- f_{- and f++ f- for individual values of x? Try looking at specific f and g functions. Suppose f(x)= 2x, g(x)= x. What are f+ and f-?}

 

[jdz86]

yep, definitely wrote it wrong, (a) was what was given, thought it was a question.

the question was something like this: using what was given, graph each of the following on the given axis, f(x),g(x), f \wedge
g, f \vee g:
f(x)=sinx, g(x)=cosx, x in [0,2pi] and graph f(x)=x(x-1)(x-2)(x-3), g(x)=0, x in [0,3]

and then (b) above was correct, using what was defined show that