Free Printable Function Operations Worksheets for Students - Free Printable

Problem: Function Operations

The task involves performing various operations on functions, such as addition, subtraction, multiplication, and composition. Below, I will solve each problem step by step.

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#### 1) \( g(n) = n^2 + 4 + 2n \), \( h(n) = -3n + 2 \)
Find \( (g \cdot h)(1) \).

Solution:
First, compute \( g(1) \):
\[
g(n) = n^2 + 4 + 2n
\]
\[
g(1) = 1^2 + 4 + 2(1) = 1 + 4 + 2 = 7
\]

Next, compute \( h(1) \):
\[
h(n) = -3n + 2
\]
\[
h(1) = -3(1) + 2 = -3 + 2 = -1
\]

Now, find \( (g \cdot h)(1) \):
\[
(g \cdot h)(1) = g(1) \cdot h(1) = 7 \cdot (-1) = -7
\]

Answer:
\[
\boxed{-7}
\]

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#### 2) \( f(x) = 4x - 3 \), \( g(x) = x^3 + 2x \)
Find \( (f - g)(4) \).

Solution:
First, compute \( f(4) \):
\[
f(x) = 4x - 3
\]
\[
f(4) = 4(4) - 3 = 16 - 3 = 13
\]

Next, compute \( g(4) \):
\[
g(x) = x^3 + 2x
\]
\[
g(4) = 4^3 + 2(4) = 64 + 8 = 72
\]

Now, find \( (f - g)(4) \):
\[
(f - g)(4) = f(4) - g(4) = 13 - 72 = -59
\]

Answer:
\[
\boxed{-59}
\]

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#### 3) \( h(x) = 3x + 3 \), \( g(x) = -4x + 1 \)
Find \( (h + g)(10) \).

Solution:
First, compute \( h(10) \):
\[
h(x) = 3x + 3
\]
\[
h(10) = 3(10) + 3 = 30 + 3 = 33
\]

Next, compute \( g(10) \):
\[
g(x) = -4x + 1
\]
\[
g(10) = -4(10) + 1 = -40 + 1 = -39
\]

Now, find \( (h + g)(10) \):
\[
(h + g)(10) = h(10) + g(10) = 33 + (-39) = 33 - 39 = -6
\]

Answer:
\[
\boxed{-6}
\]

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#### 4) \( g(a) = 3a + 2 \), \( f(a) = 2a - 4 \)
Find \( \left( \frac{g}{f} \right)(3) \).

Solution:
First, compute \( g(3) \):
\[
g(a) = 3a + 2
\]
\[
g(3) = 3(3) + 2 = 9 + 2 = 11
\]

Next, compute \( f(3) \):
\[
f(a) = 2a - 4
\]
\[
f(3) = 2(3) - 4 = 6 - 4 = 2
\]

Now, find \( \left( \frac{g}{f} \right)(3) \):
\[
\left( \frac{g}{f} \right)(3) = \frac{g(3)}{f(3)} = \frac{11}{2}
\]

Answer:
\[
\boxed{\frac{11}{2}}
\]

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#### 5) \( g(x) = 2x - 5 \), \( h(x) = 4x + 5 \)
Find \( g(3) - h(3) \).

Solution:
First, compute \( g(3) \):
\[
g(x) = 2x - 5
\]
\[
g(3) = 2(3) - 5 = 6 - 5 = 1
\]

Next, compute \( h(3) \):
\[
h(x) = 4x + 5
\]
\[
h(3) = 4(3) + 5 = 12 + 5 = 17
\]

Now, find \( g(3) - h(3) \):
\[
g(3) - h(3) = 1 - 17 = -16
\]

Answer:
\[
\boxed{-16}
\]

---

#### 6) \( g(a) = 2a - 1 \), \( h(a) = 3a - 3 \)
Find \( (g \cdot h)(-4) \).

Solution:
First, compute \( g(-4) \):
\[
g(a) = 2a - 1
\]
\[
g(-4) = 2(-4) - 1 = -8 - 1 = -9
\]

Next, compute \( h(-4) \):
\[
h(a) = 3a - 3
\]
\[
h(-4) = 3(-4) - 3 = -12 - 3 = -15
\]

Now, find \( (g \cdot h)(-4) \):
\[
(g \cdot h)(-4) = g(-4) \cdot h(-4) = (-9) \cdot (-15) = 135
\]

Answer:
\[
\boxed{135}
\]

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#### 7) \( g(t) = t^2 + 3 \), \( h(t) = 4t - 3 \)
Find \( (g \cdot h)(-1) \).

Solution:
First, compute \( g(-1) \):
\[
g(t) = t^2 + 3
\]
\[
g(-1) = (-1)^2 + 3 = 1 + 3 = 4
\]

Next, compute \( h(-1) \):
\[
h(t) = 4t - 3
\]
\[
h(-1) = 4(-1) - 3 = -4 - 3 = -7
\]

Now, find \( (g \cdot h)(-1) \):
\[
(g \cdot h)(-1) = g(-1) \cdot h(-1) = 4 \cdot (-7) = -28
\]

Answer:
\[
\boxed{-28}
\]

---

#### 8) \( g(n) = 3n + 2 \), \( f(n) = 2n^2 + 5 \)
Find \( g(f(2)) \).

Solution:
First, compute \( f(2) \):
\[
f(n) = 2n^2 + 5
\]
\[
f(2) = 2(2)^2 + 5 = 2(4) + 5 = 8 + 5 = 13
\]

Next, compute \( g(f(2)) = g(13) \):
\[
g(n) = 3n + 2
\]
\[
g(13) = 3(13) + 2 = 39 + 2 = 41
\]

Answer:
\[
\boxed{41}
\]

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#### 9) \( g(x) = -x^2 - 1 - 2x \), \( f(x) = x + 5 \)
Find \( (g - f)(x) \).

Solution:
Compute \( (g - f)(x) \):
\[
(g - f)(x) = g(x) - f(x)
\]
\[
g(x) = -x^2 - 1 - 2x
\]
\[
f(x) = x + 5
\]
\[
(g - f)(x) = (-x^2 - 1 - 2x) - (x + 5)
\]
\[
(g - f)(x) = -x^2 - 1 - 2x - x - 5
\]
\[
(g - f)(x) = -x^2 - 3x - 6
\]

Answer:
\[
\boxed{-x^2 - 3x - 6}
\]

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#### 10) \( f(x) = 3x - 1 \), \( g(x) = x^2 - x \)
Find \( \left( \frac{f}{g} \right)(x) \).

Solution:
Compute \( \left( \frac{f}{g} \right)(x) \):
\[
\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}
\]
\[
f(x) = 3x - 1
\]
\[
g(x) = x^2 - x
\]
\[
\left( \frac{f}{g} \right)(x) = \frac{3x - 1}{x^2 - x}
\]

Answer:
\[
\boxed{\frac{3x - 1}{x^2 - x}}
\]

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#### 11) \( g(a) = -3a - 3 \), \( f(a) = a^2 + 5 \)
Find \( (g - f)(a) \).

Solution:
Compute \( (g - f)(a) \):
\[
(g - f)(a) = g(a) - f(a)
\]
\[
g(a) = -3a - 3
\]
\[
f(a) = a^2 + 5
\]
\[
(g - f)(a) = (-3a - 3) - (a^2 + 5)
\]
\[
(g - f)(a) = -3a - 3 - a^2 - 5
\]
\[
(g - f)(a) = -a^2 - 3a - 8
\]

Answer:
\[
\boxed{-a^2 - 3a - 8}
\]

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#### 12) \( h(t) = 2t + 1 \), \( g(t) = 2t + 2 \)
Find \( (h - g)(t) \).

Solution:
Compute \( (h - g)(t) \):
\[
(h - g)(t) = h(t) - g(t)
\]
\[
h(t) = 2t + 1
\]
\[
g(t) = 2t + 2
\]
\[
(h - g)(t) = (2t + 1) - (2t + 2)
\]
\[
(h - g)(t) = 2t + 1 - 2t - 2
\]
\[
(h - g)(t) = -1
\]

Answer:
\[
\boxed{-1}
\]

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Final Answers:

\[
\boxed{-7, -59, -6, \frac{11}{2}, -16, 135, -28, 41, -x^2 - 3x - 6, \frac{3x - 1}{x^2 - x}, -a^2 - 3a - 8, -1}
\]