Unformatted text preview: as usual and get p−1 (x) = 5003 2x . The domain
−1 should match the range of p, which is [1, 250], and as such, we restrict the domain of
p−1 to 1 ≤ x ≤ 250.
2. We ﬁnd p−1 (220) = 500−3
= 20. Since the function p took as inputs the weekly sales and
furnished the price per system as the output, p−1 takes the price per system and returns the
weekly sales as its output. Hence, p−1 (220) = 20 means 20 systems will be sold in a week if
the price is set at $220 per system.
3. We compute P ◦ p−1 (x) = P p−1 (x) = P 5003 2x = −1.5 5003 2x + 170 5003 2x − 150.
7 we obtain P ◦ p−1 (x) = − 2 x2 +220x − 40450 .
After a hefty amount of Elementary Algebra,
To understand what this means, recall that the original proﬁt function P gave us the weekly
proﬁt as a function of the weekly sales. The function p−1 gives us the weekly sales as a
function of the price. Hence, P ◦ p−1 takes as its input a price. The function p−1 returns the
weekly sales, which in turn is fed into P to return t...
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