Proof Inequality: g(x) >= f(x) on S of Real Numbers

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Homework Statement

If f(x) = (x-1)^2 and g(x) = x+1, then g is greater than or equal to f on the set S = {real numbers x : x is between 0 and 3}.

Homework Equations

g is greater than or equal to f on the set S of real numbers iff for all s in S, g(s) is greater than f(s).

The Attempt at a Solution

Since we know x is an element of S, we know that x is between 0 and 3. That is, (x)(x-3) is less than or equal to 0. And here is where I get stuck.

I have tons of scratch paper that doesn't really show anything, and my TA gave what he calls a "proof" of this, but he assumed f(x) is less than or equal to g(x), but doesn't realize that assuming what you are trying to prove is not a way to prove anything. I just don't know how to start this.

 

[Mark44]

Hopefully, you have sketched a graph of both functions. If so, you should see that the two curves intersect at (0, 1) and (3, 4).

Look at the expression x + 1 - (x - 1)². On any interval where this expression is positive, g(x) > f(x). On any interval where the expression is negative, g(x) < f(x). Note that I am not a priori assuming either function is larger than the other.

 

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For some reason I never thought to subtract f(x) from g(x). It is proved! Thank you!