Algebra 2 Inverse Functions Worksheet - Free Printable

Let's go through the problem step by step and verify each of the answers you've written. The key concept here is function composition, denoted as $(f \circ g)(x)$, which means $f(g(x))$: you plug $g(x)$ into $f(x)$.

Given functions:
- $ f(x) = x^2 $
- $ g(x) = x - 1 $
- $ h(x) = 2x + 5 $
- $ j(x) = 6x $
- $ k(x) = x^2 + 2x + 1 $

We'll evaluate each expression carefully.

---

1) $ (j \circ h)\left(\frac{1}{2}\right) $

This means: $ j(h\left(\frac{1}{2}\right)) $

First, compute $ h\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 5 = 1 + 5 = 6 $

Then, $ j(6) = 6 \cdot 6 = 36 $

✔ Your answer: 36 — ✔️ Correct

---

2) $ g(k(4)) $

First, $ k(4) = (4)^2 + 2(4) + 1 = 16 + 8 + 1 = 25 $

Then, $ g(25) = 25 - 1 = 24 $

✔ Your answer: 24 — ✔️ Correct

---

3) $ (h \circ f)(-5) $

This is $ h(f(-5)) $

$ f(-5) = (-5)^2 = 25 $

Then $ h(25) = 2(25) + 5 = 50 + 5 = 55 $

✘ Your answer: You wrote $ (h \circ f)(-5) $ but computed $ h(-5) $? Wait — you didn't write anything for this one in red. But your handwritten work shows:

> $ (h \circ f)(-5) = ? $
> You wrote: $ (h \circ f)(-5) $, then nothing?

Wait — actually, looking at your paper, it seems you left this blank or skipped it.

But based on the above, correct value is:

✔ $ (h \circ f)(-5) = h(f(-5)) = h(25) = 50 + 5 = 55 $

So you missed this one.

---

4) $ k(h(-1)) $

First, $ h(-1) = 2(-1) + 5 = -2 + 5 = 3 $

Then $ k(3) = (3)^2 + 2(3) + 1 = 9 + 6 + 1 = 16 $

✔ Your answer: 16 — ✔️ Correct

---

5) $ g(j(h(4))) $

We do innermost first.

$ h(4) = 2(4) + 5 = 8 + 5 = 13 $

$ j(13) = 6 \cdot 13 = 78 $

$ g(78) = 78 - 1 = 77 $

✔ Your answer: 77 — ✔️ Correct

---

6) $ (h \circ g)(x) $

This is $ h(g(x)) $

$ g(x) = x - 1 $

So $ h(x - 1) = 2(x - 1) + 5 = 2x - 2 + 5 = 2x + 3 $

✔ Your answer: $ 2x + 3 $ — ✔️ Correct

---

7) $ (f \circ h)(x) $

$ f(h(x)) = f(2x + 5) = (2x + 5)^2 = 4x^2 + 20x + 25 $

✔ Your answer: $ 4x^2 + 20x + 25 $ — ✔️ Correct

---

8) $ (j \circ h)(x) $

$ j(h(x)) = j(2x + 5) = 6(2x + 5) = 12x + 30 $

✔ Your answer: $ 12x + 30 $ — ✔️ Correct

---

9) $ (k \circ j)(x) $

$ k(j(x)) = k(6x) = (6x)^2 + 2(6x) + 1 = 36x^2 + 12x + 1 $

✔ Your answer: $ 36x^2 + 12x + 1 $ — ✔️ Correct

---

10) $ (h \circ j)(x) $

$ h(j(x)) = h(6x) = 2(6x) + 5 = 12x + 5 $

✔ Your answer: $ 12x + 5 $ — ✔️ Correct

---

11) $ (f \circ g)(x) $

$ f(g(x)) = f(x - 1) = (x - 1)^2 = x^2 - 2x + 1 $

✔ Your answer: $ x^2 - 2x + 1 $ — ✔️ Correct

---

12) $ (h \circ h)(x) $

$ h(h(x)) = h(2x + 5) = 2(2x + 5) + 5 = 4x + 10 + 5 = 4x + 15 $

✔ Your answer: $ 4x + 15 $ — ✔️ Correct

---

13) $ (g \circ f)(x) $

$ g(f(x)) = g(x^2) = x^2 - 1 $

✔ Your answer: $ x^2 - 1 $ — ✔️ Correct

---

✔ Summary:

You got all correct except #3, which you either skipped or didn't write.

Let’s fix #3:

---

3) $ (h \circ f)(-5) $

As shown:
- $ f(-5) = (-5)^2 = 25 $
- $ h(25) = 2(25) + 5 = 50 + 5 = 55 $

✔ So the answer is 55

---

Final Answers:

1) 36
2) 24
3) 55 ← (you missed this)
4) 16
5) 77
6) $ 2x + 3 $
7) $ 4x^2 + 20x + 25 $
8) $ 12x + 30 $
9) $ 36x^2 + 12x + 1 $
10) $ 12x + 5 $
11) $ x^2 - 2x + 1 $
12) $ 4x + 15 $
13) $ x^2 - 1 $

All others are correct! Great job!

Just remember: order matters in function composition — always apply the inner function first.