Question

1. (2 points) let f and g be functions defined on the set of all real numbers.

a. is the statement true or false?
if f(x) > g(x) for all x, then f(x) > g(x) for all x.
b. if the statement is true, explain how you know. if the statement is false, give an example.

Explanation:

Step1: Consider the difference function

Let \(h(x)=f(x)-g(x)\). Then \(h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)\). Given \(f^{\prime}(x)>g^{\prime}(x)\) for all \(x\), so \(h^{\prime}(x)>0\) for all \(x\). This means \(h(x)\) is an increasing function.

Step2: Analyze the relationship between \(h(x)\) values

Just because \(h(x)\) is increasing does not mean \(h(x)>0\) for all \(x\). For example, if \(h(x)=x - 2\), \(h^{\prime}(x)=1>0\) (so \(h(x)\) is increasing), but when \(x = 1\), \(h(1)=1 - 2=-1<0\).

Answer:

a. False.
b. As shown in the example above, even if the derivative of \(h(x)=f(x)-g(x)\) (i.e., \(h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)\)) is positive (so \(h(x)\) is increasing), \(h(x)\) (and thus \(f(x)-g(x)\)) can be negative for some \(x\). For instance, let \(f(x)=x\) and \(g(x)=2\). Then \(f^{\prime}(x) = 1\), \(g^{\prime}(x)=0\), so \(f^{\prime}(x)>g^{\prime}(x)\) for all \(x\). But when \(x = 1\), \(f(1)=1\) and \(g(1)=2\), and \(f(1)<g(1)\).