Compose two functions both ways, (f∘g)(x) and (g∘f)(x), or evaluate a composition at a point. 59 problems with worked solutions; in 59 of them every
equation is proved by a computer algebra system.
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Let \( \displaystyle f(x) = x^{2} - 5 \) and \( \displaystyle g(x) = 3 x + 1 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{2} - 5 \right|_{\substack{ x=3 x + 1 }} = \left(3 x + 1\right)^{2} - 5 \]
f(g(x)), simplified.✓ Proved
\[ \left. 3 x + 1 \right|_{\substack{ x=x^{2} - 5 }} = 3 x^{2} - 14 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \left(3 x + 1\right)^{2} - 5,\quad (g \circ f)(x) = 3 x^{2} - 14 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{3} - 2 \) and \( \displaystyle g(x) = \frac{1}{x - 2} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{3} - 2 \right|_{\substack{ x=1 \frac{1}{x - 2} }} = -2 + \frac{1}{\left(x - 2\right)^{3}} \]
f(g(x)), simplified.✓ Proved
\[ \left. 1 \frac{1}{x - 2} \right|_{\substack{ x=x^{3} - 2 }} = \frac{1}{x^{3} - 4} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = -2 + \frac{1}{\left(x - 2\right)^{3}},\quad (g \circ f)(x) = \frac{1}{x^{3} - 4} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = 4 x + 5 \) and \( \displaystyle g(x) = \frac{1}{x - 1} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 4 x + 5 \right|_{\substack{ x=1 \frac{1}{x - 1} }} = \frac{5 x - 1}{x - 1} \]
f(g(x)), simplified.✓ Proved
\[ \left. 1 \frac{1}{x - 1} \right|_{\substack{ x=4 x + 5 }} = \frac{1}{4 x + 4} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \frac{5 x - 1}{x - 1},\quad (g \circ f)(x) = \frac{1}{4 \left(x + 1\right)} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{2} - 3 \) and \( \displaystyle g(x) = 2 x^{2} - x \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{2} - 3 \right|_{\substack{ x=2 x^{2} - x }} = x^{2} \left(2 x - 1\right)^{2} - 3 \]
f(g(x)), simplified.✓ Proved
\[ \left. 2 x^{2} - x \right|_{\substack{ x=x^{2} - 3 }} = 2 x^{4} - 13 x^{2} + 21 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = x^{2} \left(2 x - 1\right)^{2} - 3,\quad (g \circ f)(x) = 2 x^{4} - 13 x^{2} + 21 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = \sqrt{x + 1} \) and \( \displaystyle g(x) = x + 1 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. \sqrt{x + 1} \right|_{\substack{ x=x + 1 }} = \sqrt{x + 2} \]
f(g(x)), simplified.✓ Proved
\[ \left. x + 1 \right|_{\substack{ x=\sqrt{x + 1} }} = \sqrt{x + 1} + 1 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \sqrt{x + 2},\quad (g \circ f)(x) = \sqrt{x + 1} + 1 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = \frac{1}{x + 3} \) and \( \displaystyle g(x) = 2 x + 2 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 1 \frac{1}{x + 3} \right|_{\substack{ x=2 x + 2 }} = \frac{1}{2 x + 5} \]
f(g(x)), simplified.✓ Proved
\[ \left. 2 x + 2 \right|_{\substack{ x=1 \frac{1}{x + 3} }} = \frac{2 x + 8}{x + 3} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \frac{1}{2 x + 5},\quad (g \circ f)(x) = \frac{2 \left(x + 4\right)}{x + 3} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = 4 - 2 x \) and \( \displaystyle g(x) = x^{2} + 5 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 4 - 2 x \right|_{\substack{ x=x^{2} + 5 }} = - 2 x^{2} - 6 \]
f(g(x)), simplified.✓ Proved
\[ \left. x^{2} + 5 \right|_{\substack{ x=4 - 2 x }} = 4 \left(x - 2\right)^{2} + 5 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = - 2 x^{2} - 6,\quad (g \circ f)(x) = 4 \left(x - 2\right)^{2} + 5 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{2} - 5 \) and \( \displaystyle g(x) = \frac{1}{x + 1} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{2} - 5 \right|_{\substack{ x=1 \frac{1}{x + 1} }} = -5 + \frac{1}{\left(x + 1\right)^{2}} \]
f(g(x)), simplified.✓ Proved
\[ \left. 1 \frac{1}{x + 1} \right|_{\substack{ x=x^{2} - 5 }} = \frac{1}{x^{2} - 4} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = -5 + \frac{1}{\left(x + 1\right)^{2}},\quad (g \circ f)(x) = \frac{1}{x^{2} - 4} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{3} \) and \( \displaystyle g(x) = 3 x + 1 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{3} \right|_{\substack{ x=3 x + 1 }} = \left(3 x + 1\right)^{3} \]
f(g(x)), simplified.✓ Proved
\[ \left. 3 x + 1 \right|_{\substack{ x=x^{3} }} = 3 x^{3} + 1 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \left(3 x + 1\right)^{3},\quad (g \circ f)(x) = 3 x^{3} + 1 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = 5 - 2 x \) and \( \displaystyle g(x) = - 2 x^{2} - 4 x \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 5 - 2 x \right|_{\substack{ x=- 2 x^{2} - 4 x }} = 4 x^{2} + 8 x + 5 \]
f(g(x)), simplified.✓ Proved
\[ \left. - 2 x^{2} - 4 x \right|_{\substack{ x=5 - 2 x }} = - 8 x^{2} + 48 x - 70 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = 4 x^{2} + 8 x + 5,\quad (g \circ f)(x) = - 8 x^{2} + 48 x - 70 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = - 2 x^{2} - 4 x \) and \( \displaystyle g(x) = x^{2} + 3 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. - 2 x^{2} - 4 x \right|_{\substack{ x=x^{2} + 3 }} = - 2 x^{4} - 16 x^{2} - 30 \]
f(g(x)), simplified.✓ Proved
\[ \left. x^{2} + 3 \right|_{\substack{ x=- 2 x^{2} - 4 x }} = 4 x^{2} \left(x + 2\right)^{2} + 3 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = - 2 x^{4} - 16 x^{2} - 30,\quad (g \circ f)(x) = 4 x^{2} \left(x + 2\right)^{2} + 3 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = 3 x^{2} - 5 x \) and \( \displaystyle g(x) = \frac{1}{x + 2} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 3 x^{2} - 5 x \right|_{\substack{ x=1 \frac{1}{x + 2} }} = \frac{- 5 x - 7}{x^{2} + 4 x + 4} \]
f(g(x)), simplified.✓ Proved
\[ \left. 1 \frac{1}{x + 2} \right|_{\substack{ x=3 x^{2} - 5 x }} = \frac{1}{3 x^{2} - 5 x + 2} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \frac{- 5 x - 7}{x^{2} + 4 x + 4},\quad (g \circ f)(x) = \frac{1}{3 x^{2} - 5 x + 2} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = \frac{1}{x - 2} \) and \( \displaystyle g(x) = - 3 x^{2} - 3 x \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 1 \frac{1}{x - 2} \right|_{\substack{ x=- 3 x^{2} - 3 x }} = - \frac{1}{3 x^{2} + 3 x + 2} \]
f(g(x)), simplified.✓ Proved
\[ \left. - 3 x^{2} - 3 x \right|_{\substack{ x=1 \frac{1}{x - 2} }} = \frac{3 - 3 x}{x^{2} - 4 x + 4} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = - \frac{1}{3 x^{2} + 3 x + 2},\quad (g \circ f)(x) = \frac{3 \left(1 - x\right)}{x^{2} - 4 x + 4} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x + 1 \) and \( \displaystyle g(x) = x^{2} - 4 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x + 1 \right|_{\substack{ x=x^{2} - 4 }} = x^{2} - 3 \]
f(g(x)), simplified.✓ Proved
\[ \left. x^{2} - 4 \right|_{\substack{ x=x + 1 }} = \left(x + 1\right)^{2} - 4 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = x^{2} - 3,\quad (g \circ f)(x) = \left(x + 1\right)^{2} - 4 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x + 1 \) and \( \displaystyle g(x) = \frac{1}{x + 2} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x + 1 \right|_{\substack{ x=1 \frac{1}{x + 2} }} = \frac{x + 3}{x + 2} \]
f(g(x)), simplified.✓ Proved
\[ \left. 1 \frac{1}{x + 2} \right|_{\substack{ x=x + 1 }} = \frac{1}{x + 3} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \frac{x + 3}{x + 2},\quad (g \circ f)(x) = \frac{1}{x + 3} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = - 3 x^{2} + 3 x \) and \( \displaystyle g(x) = x^{2} - 3 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. - 3 x^{2} + 3 x \right|_{\substack{ x=x^{2} - 3 }} = - 3 x^{4} + 21 x^{2} - 36 \]
f(g(x)), simplified.✓ Proved
\[ \left. x^{2} - 3 \right|_{\substack{ x=- 3 x^{2} + 3 x }} = 9 x^{2} \left(x - 1\right)^{2} - 3 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = - 3 x^{4} + 21 x^{2} - 36,\quad (g \circ f)(x) = 9 x^{2} \left(x - 1\right)^{2} - 3 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = 5 - x \) and \( \displaystyle g(x) = x^{2} + 2 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. 5 - x \right|_{\substack{ x=x^{2} + 2 }} = 3 - x^{2} \]
f(g(x)), simplified.✓ Proved
\[ \left. x^{2} + 2 \right|_{\substack{ x=5 - x }} = \left(x - 5\right)^{2} + 2 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = 3 - x^{2},\quad (g \circ f)(x) = \left(x - 5\right)^{2} + 2 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = - 3 x - 4 \) and \( \displaystyle g(x) = 3 x + 1 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. - 3 x - 4 \right|_{\substack{ x=3 x + 1 }} = - 9 x - 7 \]
f(g(x)), simplified.✓ Proved
\[ \left. 3 x + 1 \right|_{\substack{ x=- 3 x - 4 }} = - 9 x - 11 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = - 9 x - 7,\quad (g \circ f)(x) = - 9 x - 11 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{3} - 1 \) and \( \displaystyle g(x) = \sqrt{x + 5} \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{3} - 1 \right|_{\substack{ x=\sqrt{x + 5} }} = \left(x + 5\right)^{\frac{3}{2}} - 1 \]
f(g(x)), simplified.✓ Proved
\[ \left. \sqrt{x + 5} \right|_{\substack{ x=x^{3} - 1 }} = \sqrt{x^{3} + 4} \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \left(x + 5\right)^{\frac{3}{2}} - 1,\quad (g \circ f)(x) = \sqrt{x^{3} + 4} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Let \( \displaystyle f(x) = x^{3} + 5 \) and \( \displaystyle g(x) = 2 x + 1 \). Find \( \displaystyle (f \circ g)(x) \) and \( \displaystyle (g \circ f)(x) \), simplified.
Show the first step
Show the whole solution
(f∘g)(x) = f(g(x)): put g(x) in place of x in f. (g∘f) is the other way round.
Reviewed
\[ \left. x^{3} + 5 \right|_{\substack{ x=2 x + 1 }} = \left(2 x + 1\right)^{3} + 5 \]
f(g(x)), simplified.✓ Proved
\[ \left. 2 x + 1 \right|_{\substack{ x=x^{3} + 5 }} = 2 x^{3} + 11 \]
g(f(x)), simplified.✓ Proved
The two orders give different functions: composition is not commutative.
Reviewed
Answer \( (f \circ g)(x) = \left(2 x + 1\right)^{3} + 5,\quad (g \circ f)(x) = 2 x^{3} + 11 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify