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Evaluate................

Evaluate................-example-1
Evaluate................-example-1
Evaluate................-example-2
User Ernewston
by
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1 Answer

4 votes

Answer:

h(8q²-2q) = 56q² -10q

k(2q²+3q) = 16q² +31q

Explanation:

1. Replace x in the function definition with the function's argument, then simplify.

h(x) = 7x +4q

h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q

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2. Same as the first problem.

k(x) = 8x +7q

k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q

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Comment on the problem

In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...

h(8q² -2q) = 56q² -14q +4z

and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.

In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.

User Grundic
by
8.4k points

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