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Problem : What is x2 + 3x - 5?

x2 + 3x - 5 = 5 (direct substitution).

Problem : What is ?

= 0. By factoring x out of the numerator and denominator, f is defined at x = 0, and direct substitution works.

Problem : What is 4?

4 = 4. This is a constant function.

Problem : What is ?

By factoring (x - 3) out of the numerator and the denominator, we find = = .

Problem : Let f (x) = x for x < 1, f (x) = 1 for x≥1. What is f (x)?

f (x) = 1.

Problem : Let f (x) = x2 for x < 0, f (x) = - 1 for x≥ 0. What are f (x), f (x), and f (x)?

f (x) = 0. f (x) = - 1. f (x) doesn't exist, because f (x)≠f (x).
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