Unit Test — Cumulative exam Answers

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Given . What is ?

Negative five-thirds

Negative four-thirds

Four-thirds

Five-thirds

Consider the piecewise function.

Limit of f (x) = 2 as x approaches 1 minus. Limit of f (x) = 9 as x approaches 1 plus.

Limit of f (x) = 9 as x approaches 1 minus. Limit of f (x) = 2 as x approaches 1 plus.

Limit of f (x) = negative 4 as x approaches 1 minus. Limit of f (x) = negative 4 as x approaches 1 plus.

Limit of f (x) D N E as x approaches 1 minus. Limit of f (x) D N E as x approaches 1 plus.

Review the table of values for function h(x).xh(x)8.93.838.99–1.198.9994.739undefined9.001–4.739.011.199.1–3.83

The limit does not exist because the values of h(x) seem to oscillate between random values around x = 9.

The limit does exist because h(x) is defined for all given values around x = 9, even though h(x) isn’t defined at x = 9.

The limit does not exist because h(x) is not defined at x = 9, and for a limit to exist, the function must be defined at x = 9.

The limit does exist because the values of h(x) to the right of x = 9 are all opposites of the values of h(x) to the left of x = 9.

Review the graph of function f(x), which is defined for –6 ≤ x ≤ 8.

Limit of f (x) as x approaches 2 plus

Limit of f (x) as x approaches 2 minus

Limit of f (x) as x approaches negative 6 plus

Limit of f (x) as x approaches negative 6 minus

Given f(x) = 3x2 + x and g(x) = 2x. What is ?

104

188

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