Question

Use the definitions of increasing and decreasing functions to prove that \(f(x)=1 / x\) is decreasing on \((0, \infty)\).

Short answer

By choosing any two arbitrary points 'a' and 'b' within the domain (0, \(\infty\)) and such that \(a < b\), and by evaluating the function at these points, it can be observed that \(f(a) > f(b)\). Hence, by definition, the function \(f(x) = 1 / x\) is decreasing for \(x > 0\).

Step-by-step solution

Define the function and its domain

The function to be tested is \(f(x) = 1 / x\) and the domain that we need to consider is \(x\) greater than 0 (i.e., (0, \(\infty\))).

Choose arbitrary numbers within the domain

Select arbitrary numbers \'a\' and \'b\' in the domain such that \(0 < a < b\). Now, the task is to prove that whenever \(a < b\), \(f(a) > f(b)\). Let's compute the function values at these points and compare them.

Compute function values at chosen points

The value of the function \(f(x)\) at points \'a\' and \'b\' are \(f(a) = 1/a\) and \(f(b) = 1/b\) respectively.

Comparison of function values

Since a and b are positive and \(a < b\), it implies that \(1/a > 1/b\). Thus, it can be said that \(f(a) > f(b)\) whenever \(a < b\). Therefore, by definition, the function \(f(x) = 1 / x\) is decreasing on the domain (0, \(\infty\)).