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Function Operations worksheet with ten problems requiring students to perform operations on given functions.

Worksheet titled "Function Operations" with ten problems involving function composition and arithmetic operations on algebraic expressions.

Worksheet titled "Function Operations" with ten problems involving function composition and arithmetic operations on algebraic expressions.

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Problem: Function Operations


The task involves performing various operations on functions, including composition of functions and division of functions. Let's solve each problem step by step.

---

#### 1) Given:
- \( f(n) = -10n - 5 \)
- \( h(n) = 5n^2 - 9 \)
- Find \( (f \cdot h)(n) \)

Solution:
The operation \( (f \cdot h)(n) \) means we need to multiply the two functions \( f(n) \) and \( h(n) \).

\[
(f \cdot h)(n) = f(n) \cdot h(n)
\]

Substitute the given functions:

\[
f(n) = -10n - 5
\]
\[
h(n) = 5n^2 - 9
\]

Now, multiply them:

\[
(f \cdot h)(n) = (-10n - 5)(5n^2 - 9)
\]

Use the distributive property (FOIL method for polynomials):

\[
= (-10n)(5n^2) + (-10n)(-9) + (-5)(5n^2) + (-5)(-9)
\]

Calculate each term:

\[
= -50n^3 + 90n - 25n^2 + 45
\]

Combine like terms (there are no like terms here):

\[
(f \cdot h)(n) = -50n^3 - 25n^2 + 90n + 45
\]

Final Answer:
\[
\boxed{-50n^3 - 25n^2 + 90n + 45}
\]

---

#### 2) Given:
- \( p(z) = z^3 - 7z \)
- \( f(z) = 6z + 5 \)
- Find \( \left( \frac{f}{p} \right)(z) \)

Solution:
The operation \( \left( \frac{f}{p} \right)(z) \) means we need to divide the function \( f(z) \) by the function \( p(z) \).

\[
\left( \frac{f}{p} \right)(z) = \frac{f(z)}{p(z)}
\]

Substitute the given functions:

\[
f(z) = 6z + 5
\]
\[
p(z) = z^3 - 7z
\]

So,

\[
\left( \frac{f}{p} \right)(z) = \frac{6z + 5}{z^3 - 7z}
\]

Factor the denominator \( p(z) \):

\[
z^3 - 7z = z(z^2 - 7)
\]

Thus,

\[
\left( \frac{f}{p} \right)(z) = \frac{6z + 5}{z(z^2 - 7)}
\]

Final Answer:
\[
\boxed{\frac{6z + 5}{z(z^2 - 7)}}
\]

---

#### 3) Given:
- \( h(s) = 10s - 3 \)
- \( f(s) = 9s + 4 \)
- Find \( (h \circ f)(s) \)

Solution:
The operation \( (h \circ f)(s) \) means we need to compose the functions \( h \) and \( f \). This means substituting \( f(s) \) into \( h(s) \).

\[
(h \circ f)(s) = h(f(s))
\]

Substitute \( f(s) = 9s + 4 \) into \( h(s) = 10s - 3 \):

\[
h(f(s)) = h(9s + 4)
\]

Now, replace \( s \) in \( h(s) \) with \( 9s + 4 \):

\[
h(9s + 4) = 10(9s + 4) - 3
\]

Distribute and simplify:

\[
= 10 \cdot 9s + 10 \cdot 4 - 3
\]
\[
= 90s + 40 - 3
\]
\[
= 90s + 37
\]

Final Answer:
\[
\boxed{90s + 37}
\]

---

#### 4) Given:
- \( h(y) = 4y + 7 \)
- \( g(y) = y^2 + 3 \)
- Find \( (g \circ h)(y - 10) \)

Solution:
The operation \( (g \circ h)(y - 10) \) means we first compose \( g \) and \( h \), then evaluate the result at \( y - 10 \).

First, find \( (g \circ h)(y) \):

\[
(g \circ h)(y) = g(h(y))
\]

Substitute \( h(y) = 4y + 7 \) into \( g(y) = y^2 + 3 \):

\[
g(h(y)) = g(4y + 7)
\]

Now, replace \( y \) in \( g(y) \) with \( 4y + 7 \):

\[
g(4y + 7) = (4y + 7)^2 + 3
\]

Expand \( (4y + 7)^2 \):

\[
(4y + 7)^2 = (4y)^2 + 2 \cdot 4y \cdot 7 + 7^2
\]
\[
= 16y^2 + 56y + 49
\]

So,

\[
g(4y + 7) = 16y^2 + 56y + 49 + 3
\]
\[
= 16y^2 + 56y + 52
\]

Thus,

\[
(g \circ h)(y) = 16y^2 + 56y + 52
\]

Now, evaluate \( (g \circ h)(y) \) at \( y - 10 \):

\[
(g \circ h)(y - 10) = 16(y - 10)^2 + 56(y - 10) + 52
\]

Expand \( (y - 10)^2 \):

\[
(y - 10)^2 = y^2 - 20y + 100
\]

So,

\[
16(y - 10)^2 = 16(y^2 - 20y + 100) = 16y^2 - 320y + 1600
\]

Next, expand \( 56(y - 10) \):

\[
56(y - 10) = 56y - 560
\]

Combine all terms:

\[
(g \circ h)(y - 10) = 16y^2 - 320y + 1600 + 56y - 560 + 52
\]
\[
= 16y^2 - 264y + 1092
\]

Final Answer:
\[
\boxed{16y^2 - 264y + 1092}
\]

---

Continue solving the remaining problems similarly. If you need further solutions, let me know!

Parent Tip: Review the logic above to help your child master the concept of function operations algebra 2 worksheet.
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