Composite Function Worksheet - Fill Online, Printable, Fillable ... - Free Printable

Let’s solve each problem step by step. We’ll use the given functions and plug in values or expressions as needed.

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Given for problems 1–9:
- f(x) = 2x – 1
- h(x) = x² + 4

We need to compute:

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1. f(g(-3))

Wait — we don’t have g(x) defined yet! Looking ahead, problem 10 gives us g(x), but problems 1–9 must be using a different setup? Actually, looking again — problem 4 is f(g(h(2))), which suggests that maybe g(x) was meant to be defined earlier? But it’s not.

Hold on — let me check the image again (mentally). In problem 10, they define:
> For 10–13: Let f(x) = -3x + 7 and g(x) = 2x² – 8.

But problems 1–9 say:
> Let f(x) = 2x – 1, and h(x) = x² + 4.

There’s no definition of g(x) for problems 1–9. That’s a problem.

Wait — perhaps there’s a typo in the worksheet? Or maybe g(x) is supposed to be defined somewhere else?

Looking at problem 4: f(g(h(2))) — if we can’t compute this without g(x), then maybe... actually, let me re-read the directions.

The worksheet says “For 1–9: Let f(x) = 2x – 1, and h(x) = x² + 4.” It doesn’t mention g(x). But then problems like #1, #4, #5, #6 all involve g(x).

This seems like an error in the worksheet — unless... wait! Maybe in the original context, g(x) was defined before this page? Since this is a "composite function worksheet", perhaps g(x) was given on a previous page?

But since we only have this image, and no prior info, we have to assume something.

Alternatively — look at problem 10: it defines new f and g. So maybe for 1–9, g(x) is missing? That can’t be right.

Wait — let me check problem 5: h(g(f(5))) — again needs g.

Problem 6: g(f(h(-4))) — needs g.

Unless... is it possible that “g” is a typo and should be “h”? Let’s test that idea.

Suppose in problems 1–9, wherever it says “g”, it should be “h”. Let’s try that for problem 1:

If problem 1 is f(h(-3)):

f(h(-3)) = f((-3)^2 + 4) = f(9 + 4) = f(13) = 2*13 - 1 = 26 - 1 = 25 → that works.

Problem 4: f(h(h(2)))?

h(2) = 2² + 4 = 4 + 4 = 8

h(8) = 8² + 4 = 64 + 4 = 68

f(68) = 2*68 - 1 = 136 - 1 = 135 → also works.

But problem 5: h(h(f(5)))?

f(5) = 2*5 - 1 = 10 - 1 = 9

h(9) = 81 + 4 = 85

h(85) = 85² + 4 = 7225 + 4 = 7229 → big number, but mathematically fine.

But now problem 6: h(f(h(-4)))?

h(-4) = (-4)^2 + 4 = 16 + 4 = 20

f(20) = 2*20 - 1 = 40 - 1 = 39

h(39) = 39² + 4 = 1521 + 4 = 1525 → still works.

But what about problem 2: f(h(7))? That would be fine.

Problem 3: (g∘h)(-1) — if g is h, then (h∘h)(-1) = h(h(-1))

h(-1) = 1 + 4 = 5

h(5) = 25 + 4 = 29 → okay.

But the notation (g∘h) usually means g(h(x)), so if g is not defined, we’re stuck.

Wait — another possibility: maybe in problems 1–9, “g” is actually “f”? Let’s test problem 1: f(f(-3))

f(-3) = 2*(-3) -1 = -6 -1 = -7

f(-7) = 2*(-7) -1 = -14 -1 = -15 → possible.

But problem 4: f(f(h(2))) — h(2)=8, f(8)=15, f(15)=29 → also possible.

But then problem 5: h(f(f(5))) — f(5)=9, f(9)=17, h(17)=289+4=293 → okay.

But problem 6: f(f(h(-4))) — h(-4)=20, f(20)=39, f(39)=77 → okay.

But now problem 3: (g∘h)(-1) — if g is f, then (f∘h)(-1) = f(h(-1)) = f(5) = 9 → okay.

So both interpretations are mathematically possible, but which one is intended?

Actually, looking back at the worksheet, problem 10 introduces new definitions for f and g, implying that for 1–9, f and h are defined, and g might be a mistake.

But here’s a better clue: in problem 7, 8, 9, they ask for f(x+1), g(3a), h(x-2). If g is not defined, we can’t do #8.

Unless... wait! Perhaps in the original worksheet, g(x) was defined above this section? Since this is a composite function worksheet, maybe g(x) was given on a previous page.

Since we don’t have that, and to make progress, I think there might be a typo, and for problems 1–9, when it says “g”, it should be “h”. Because otherwise, we can’t solve them.

Alternatively, maybe “g” is meant to be the same as “f”? But that doesn't make sense because they use both letters.

Another idea: perhaps in problems 1–9, “g” is a third function that wasn't printed? That would be unfair.

Let me look at problem 12 and 13 — they define f and g separately, so likely for 1–9, g is missing.

But wait — problem 4 is f(g(h(2))), and if we assume that g(x) = x^2 + 4 (same as h), then it works, as I did earlier.

Perhaps the worksheet has a formatting error, and h(x) is meant to be used where g appears in 1–9.

I think for the sake of solving, I'll assume that in problems 1–9, whenever "g" appears, it's a typo and should be "h". This allows us to compute everything with the given functions.

So let's proceed with that assumption:

For problems 1–9:
- f(x) = 2x – 1
- h(x) = x² + 4
- And wherever "g" is written, treat it as "h"

Now let's solve each:

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1. f(g(-3)) → assume f(h(-3))

h(-3) = (-3)^2 + 4 = 9 + 4 = 13
f(13) = 2*13 - 1 = 26 - 1 = 25

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2. f(h(7))

h(7) = 7^2 + 4 = 49 + 4 = 53
f(53) = 2*53 - 1 = 106 - 1 = 105

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3. (g∘h)(-1) → assume (h∘h)(-1) = h(h(-1))

h(-1) = (-1)^2 + 4 = 1 + 4 = 5
h(5) = 5^2 + 4 = 25 + 4 = 29

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4. f(g(h(2))) → f(h(h(2)))

h(2) = 2^2 + 4 = 4 + 4 = 8
h(8) = 8^2 + 4 = 64 + 4 = 68
f(68) = 2*68 - 1 = 136 - 1 = 135

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5. h(g(f(5))) → h(h(f(5)))

f(5) = 2*5 - 1 = 10 - 1 = 9
h(9) = 9^2 + 4 = 81 + 4 = 85
h(85) = 85^2 + 4 = 7225 + 4 = 7229

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6. g(f(h(-4))) → h(f(h(-4)))

h(-4) = (-4)^2 + 4 = 16 + 4 = 20
f(20) = 2*20 - 1 = 40 - 1 = 39
h(39) = 39^2 + 4 = 1521 + 4 = 1525

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7. f(x + 1)

f(x+1) = 2*(x+1) - 1 = 2x + 2 - 1 = 2x + 1

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8. g(3a) → assume h(3a)

h(3a) = (3a)^2 + 4 = 9a^2 + 4 → 9a² + 4

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9. h(x - 2)

h(x-2) = (x-2)^2 + 4 = x^2 - 4x + 4 + 4 = x² - 4x + 8

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Now for problems 10–13:

Given:
- f(x) = -3x + 7
- g(x) = 2x² – 8

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10. f(g(5))

First, g(5) = 2*(5)^2 - 8 = 2*25 - 8 = 50 - 8 = 42
Then f(42) = -3*42 + 7 = -126 + 7 = -119

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11. (g∘f)(x) = g(f(x))

f(x) = -3x + 7
g(f(x)) = 2*[f(x)]^2 - 8 = 2*(-3x + 7)^2 - 8

Compute (-3x + 7)^2 = 9x^2 - 42x + 49
Then 2*(9x^2 - 42x + 49) = 18x^2 - 84x + 98
Minus 8: 18x^2 - 84x + 90

So (g∘f)(x) = 18x² - 84x + 90

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12. If f(x) = 3x – 5 and g(x) = x³, find (f ∘ g)(3)

Note: This redefines f and g for this problem only.

(f ∘ g)(3) = f(g(3))

g(3) = 3^3 = 27
f(27) = 3*27 - 5 = 81 - 5 = 76

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13. If f(x) = -9x – 9 and g(x) = √(x – 9), find (f ∘ g)(10)

(f ∘ g)(10) = f(g(10))

g(10) = √(10 - 9) = √1 = 1
f(1) = -9*1 - 9 = -9 - 9 = -18

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Now, let’s double-check calculations for accuracy.

Recheck problem 5: h(h(f(5)))

f(5)=9, h(9)=81+4=85, h(85)=7225+4=7229 ✓

Problem 6: h(f(h(-4)))

h(-4)=16+4=20, f(20)=40-1=39, h(39)=1521+4=1525 ✓

Problem 11: g(f(x)) = 2*(-3x+7)^2 -8

(-3x+7)^2 = 9x^2 -42x +49

2*that = 18x^2 -84x +98

Minus 8: 18x^2 -84x +90 ✓

Problem 12: f(g(3)) = f(27) = 3*27-5=81-5=76 ✓

Problem 13: g(10)=√1=1, f(1)=-9-9=-18 ✓

All seem correct under our assumption that "g" in 1–9 is a typo for "h".

If the worksheet intended a different g(x), we’d need that definition, but based on available info, this is the best we can do.

Final Answers:

1. 25
2. 105
3. 29
4. 135
5. 7229
6. 1525
7. 2x + 1
8. 9a² + 4
9. x² - 4x + 8
10. -119
11. 18x² - 84x + 90
12. 76
13. -18

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Final Answer:
1. 25
2. 105
3. 29
4. 135
5. 7229
6. 1525
7. 2x + 1
8. 9a² + 4
9. x² - 4x + 8
10. -119
11. 18x² - 84x + 90
12. 76
13. -18