Composition of Functions Word Problems | PDF | Coupon | Function ... - Free Printable

Let's solve each part of this Composition of Functions – Word Problems step by step.

We are given a scenario involving two discounts:
- A $5 off coupon from the mail.
- A 25% discount at the store.

Let $ x $ represent the original cost of the jeans.

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1. Write a function, $ f(x) $, that represents the effect of your original coupon.

The coupon gives $5 off, so we subtract 5 from the original price.

$$
f(x) = x - 5
$$

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2. Write a function, $ g(x) $, that represents the effect of the 25% discount at the store.

A 25% discount means you pay 75% of the original price.

$$
g(x) = 0.75x
$$

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3. Write a function, $ h(x) $, that represents how much you would pay if you use the mail coupon first, followed by the store discount.

You apply the $5 off first, then take 25% off of the reduced price.

So:
- First: $ f(x) = x - 5 $
- Then apply the 25% discount: $ g(f(x)) = g(x - 5) $

$$
h(x) = g(f(x)) = g(x - 5) = 0.75(x - 5)
$$

$$
\boxed{h(x) = 0.75(x - 5)}
$$

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4. Write a function, $ j(x) $, that represents how much you would pay if you use the store discount first, followed by the mail coupon.

You apply the 25% discount first, then subtract $5.

So:
- First: $ g(x) = 0.75x $
- Then apply $5 off: $ f(g(x)) = f(0.75x) $

$$
j(x) = f(g(x)) = f(0.75x) = 0.75x - 5
$$

$$
\boxed{j(x) = 0.75x - 5}
$$

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5. You find a pair of jeans for $36. How much would you pay for it using both functions $ h(x) $ and $ j(x) $?

Use $ x = 36 $

#### Using $ h(x) $: Coupon first, then discount

$$
h(36) = 0.75(36 - 5) = 0.75(31) = 23.25
$$

#### Using $ j(x) $: Discount first, then coupon

$$
j(36) = 0.75(36) - 5 = 27 - 5 = 22
$$

So:
- With $ h(x) $: $23.25
- With $ j(x) $: $22.00

✔ So applying the store discount first saves more money.

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6. If you only have $40 with you, what’s the most expensive pair of jeans you can purchase? (Do not consider tax)

We want to find the maximum original price $ x $ such that after discounts, the final price is ≤ $40.

But we need to consider which order gives the best deal (i.e., lowest final price), because depending on the order, the final price differs.

From part 5, we saw that applying the 25% discount first ($ j(x) $) gives a lower final price than applying the coupon first ($ h(x) $).

So, to maximize the amount we can spend, we should use the better strategy: apply the store discount first, then the coupon → use $ j(x) = 0.75x - 5 $

Set $ j(x) \leq 40 $:

$$
0.75x - 5 \leq 40
$$

$$
0.75x \leq 45
$$

$$
x \leq \frac{45}{0.75} = 60
$$

So, the most expensive pair of jeans you can buy is $60, when using the store discount first.

Let’s verify:

- Original price: $60
- 25% off: $ 0.75 \times 60 = 45 $
- Then $5 off: $ 45 - 5 = 40 $

Perfect! You pay exactly $40.

✔ Answer: $60

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7. Determine when you would want to use $ h(x) $ (apply $5 coupon first), and when you would want to use $ j(x) $ (apply 25% off first).

Compare $ h(x) $ and $ j(x) $:

- $ h(x) = 0.75(x - 5) = 0.75x - 3.75 $
- $ j(x) = 0.75x - 5 $

Now compare:

$$
h(x) = 0.75x - 3.75 \\
j(x) = 0.75x - 5
$$

Since $ -3.75 > -5 $, we have:

$$
h(x) > j(x) \quad \text{for all } x
$$

That means: $ j(x) $ always gives a lower final price than $ h(x) $.

So, you always want to use $ j(x) $ — apply the 25% discount first, then the $5 coupon.

But wait — is this true always?

Let’s think carefully.

Actually, this depends on whether the discount applies to the full price or not.

But in our model, we assumed:
- The $5 coupon is applied only after the price has been reduced.

But what if the $5 coupon is only valid on prices above $5? Or if the store doesn't allow coupons on already discounted items?

But based on the problem statement, we assume both discounts are applicable, but order matters.

Wait — let's double-check our comparison:

We have:
- $ h(x) = 0.75(x - 5) $
- $ j(x) = 0.75x - 5 $

Now compare:

$$
h(x) - j(x) = [0.75(x - 5)] - [0.75x - 5] = 0.75x - 3.75 - 0.75x + 5 = 1.25
$$

So:
$$
h(x) = j(x) + 1.25
$$

This means $ h(x) $ is always $1.25 more than $ j(x) $, regardless of $ x $.

So, $ j(x) $ is always cheaper.

Therefore:

> ✔ You should always use $ j(x) $ — apply the 25% discount first, then the $5 coupon.

But wait — is there any situation where $ h(x) $ might be better?

Only if the $5 coupon is not valid after the discount — but the problem says "you receive a coupon for $5 off", implying it's usable.

But here’s a catch: What if the original price is less than $5?

For example, suppose $ x = 4 $

Then:
- $ h(x) = 0.75(4 - 5) = 0.75(-1) = -0.75 $ → negative price? Not possible.

But also, you can’t apply $5 off to $4 — you’d get $-1, which isn’t allowed.

So, the coupon may not be usable if the price is less than $5.

But since we're dealing with jeans priced at $36 or more, and the maximum we can afford is $60, we’re safe.

But let’s suppose the coupon is only usable if the price is ≥ $5.

Then, for very low prices, $ h(x) $ might not work.

But in general, for prices above $5, we can compare.

Let’s suppose the store allows both discounts, but the order affects the result.

But from algebra:

- $ h(x) = 0.75(x - 5) $
- $ j(x) = 0.75x - 5 $

And as shown, $ j(x) < h(x) $ for all $ x $, since $ j(x) = h(x) - 1.25 $

So, you save $1.25 more by applying the 25% discount first.

Therefore:

> ✔ Always use $ j(x) $ — apply the 25% off first, then the $5 coupon.

There is no case where $ h(x) $ is better, unless the $5 coupon cannot be applied after the discount (e.g., store policy), but the problem does not suggest that.

So, you would never want to use $ h(x) $ if both discounts are allowed.

But let’s suppose the $5 coupon is only valid on the original price, not on discounted ones.

Then, $ h(x) $ would be invalid if used after discount.

But again, the problem doesn't say that.

So based on standard assumptions:

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✔ Final Answers:

1. $ f(x) = x - 5 $

2. $ g(x) = 0.75x $

3. $ h(x) = 0.75(x - 5) $

4. $ j(x) = 0.75x - 5 $

5. For $ x = 36 $:
- $ h(36) = 0.75(31) = 23.25 $
- $ j(36) = 27 - 5 = 22 $
- Pay $23.25 with $ h(x) $, $22 with $ j(x) $

6. Most expensive jeans you can buy: $60 (using $ j(x) $)

7. You should always use $ j(x) $ (apply 25% off first, then $5 off), because it always results in a lower final price than $ h(x) $. There is no situation where $ h(x) $ is better under these conditions.

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🟩 Summary:

| Question | Answer |
|--------|--------|
| 1 | $ f(x) = x - 5 $ |
| 2 | $ g(x) = 0.75x $ |
| 3 | $ h(x) = 0.75(x - 5) $ |
| 4 | $ j(x) = 0.75x - 5 $ |
| 5 | $ h(36) = 23.25 $, $ j(36) = 22 $ |
| 6 | $60 |
| 7 | Always use $ j(x) $; $ h(x) $ is never better |

✔ Key Insight: Composition of functions matters — the order of discounts changes the outcome, and applying percentage discounts before fixed-dollar coupons usually saves more.