Problem : What is x ² + 3x - 5 ?

Solution for Problem 1 >>

x ² + 3x - 5 = 5 (direct substitution).

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Problem : What is ?

Solution for Problem 2 >>

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

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Problem : What is 4 ?

Solution for Problem 3 >>

4 = 4 . This is a constant function.

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Problem : What is ?

Solution for Problem 4 >>

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

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

Solution for Problem 5 >>

f (x) = 1 .

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Problem : Let f (x) = x ² for x < 0 , f (x) = - 1 for x≥ 0 . What are f (x) , f (x) , and f (x) ?

Solution for Problem 6 >>

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

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