Try It
ⓐ yes
ⓑ yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)
w=f(d) w=f(d)
yes
g( 5 )=1 g( 5 )=1
m=8 m=8
y=f( x )= x 3 2 y=f( x )= x 3 2
g( 1 )=8 g( 1 )=8
x=0 x=0 or x=2 x=2
ⓐ yes, because each bank account has a single balance at any given time
ⓑ no, because several bank account numbers may have the same balance
ⓒ no, because the same output may correspond to more than one input.
ⓐ Yes, letter grade is a function of percent grade;
ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.
yes
No, because it does not pass the horizontal line test.
{−5,0,5,10,15} {−5,0,5,10,15}
( −∞,∞ ) ( −∞,∞ )
( −∞, 1 2 )∪( 1 2 ,∞ ) ( −∞, 1 2 )∪( 1 2 ,∞ )
[ − 5 2 ,∞ ) [ − 5 2 ,∞ )
ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
ⓑ { x|x≤−2or−1≤x<3 } { x|x≤−2or−1≤x<3 } ;
ⓒ (−∞,−2]∪[−1,3) (−∞,−2]∪[−1,3)
domain =[1950,2002] range = [47,000,000,89,000,000]
domain: ( −∞,2 ]; ( −∞,2 ]; range: ( −∞,0 ] ( −∞,0 ]
$2.84−$2.31 5 years = $0.53 5 years =$0.106 $2.84−$2.31 5 years = $0.53 5 years =$0.106 per year.
1 2 1 2
a+7 a+7
The local maximum appears to occur at (−1,28), (−1,28), and the local minimum occurs at (5,−80). (5,−80). The function is increasing on (−∞,−1)∪(5,∞) (−∞,−1)∪(5,∞) and decreasing on (−1,5). (−1,5).
( fg )( x )=f( x )g( x )=( x−1 )( x 2 −1 )= x 3 − x 2 −x+1 ( f−g )( x )=f( x )−g( x )=( x−1 )−( x 2 −1 )=x− x 2 ( fg )( x )=f( x )g( x )=( x−1 )( x 2 −1 )= x 3 − x 2 −x+1 ( f−g )( x )=f( x )−g( x )=( x−1 )−( x 2 −1 )=x− x 2
No, the functions are not the same.
A gravitational force is still a force, so a( G(r) ) a( G(r) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G( a(F) ) G( a(F) ) does not make sense.
f(g(1))=f(3)=3 f(g(1))=f(3)=3 and g(f(4))=g(1)=3 g(f(4))=g(1)=3
g(f(2))=g(5)=3 g(f(2))=g(5)=3
ⓐ 8
ⓑ 20
[ −4,0 )∪( 0,∞ ) [ −4,0 )∪( 0,∞ )
Possible answer:
g( x )= 4+ x 2 g( x )= 4+ x 2 h( x )= 4 3−x h( x )= 4 3−x f=h∘g f=h∘g
The graphs of f(x) f(x) and g(x) g(x) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.
g( x )= 1 x-1 +1 g( x )= 1 x-1 +1
ⓐ g(x)=−f(x) g(x)=−f(x) x x -2 0 2 4 g(x) g(x) −5 −5 −10 −10 −15 −15 −20 −20
ⓑ h(x)=f(−x) h(x)=f(−x) x x -2 0 2 4 h(x) h(x) 15 10 5 unknown
Notice: g(x)=f(−x) g(x)=f(−x) looks the same as f(x) f(x).
even
row: x x | 2 | 4 | 6 | 8
row: g(x) g(x) | 9 | 12 | 15 | 0
g(x)=3x-2
g(x)=f( 1 3 x ) ਇਸ ਲਈ ਵਰਗਮੂਲ ਫੰਕਸ਼ਨ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋਏ ਸਾਨੂੰ g(x)= 1 3 x ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ।
| x−2 |≤3
p ਵੇਰੀਏਬਲ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋਏ, ਪਾਸ ਕਰਨ ਲਈ, | p−80 |≤20
f(x)=−| x+2 |+3
x=−1 ਜਾਂ x=2
f(0)=1, ਇਸ ਲਈ ਗ੍ਰਾਫ਼ y-ਧੁਰੇ ਨੂੰ (0,1) 'ਤੇ ਕੱਟਦਾ ਹੈ। f(x)=0 ਜਦੋਂ x=−5 ਅਤੇ x=1 ਹੁੰਦਾ ਹੈ, ਇਸ ਲਈ ਗ੍ਰਾਫ਼ x-ਧੁਰੇ ਨੂੰ (−5,0) ਅਤੇ (1,0) 'ਤੇ ਕੱਟਦਾ ਹੈ।
-8≤x≤4
k≤1 ਜਾਂ k≥7; ਅੰਤਰਾਲ ਨੋਟੇਸ਼ਨ ਵਿੱਚ, ਇਹ (−∞,1]∪[7,∞) ਹੋਵੇਗਾ।
h(2)=6
ਹਾਂ
ਹਾਂ
ਫੰਕਸ਼ਨ f −1 ਦਾ ਡੋਮੇਨ (−∞,−2) ਹੈ ਅਤੇ ਫੰਕਸ਼ਨ f −1 ਦੀ ਰੇਂਜ (1,∞) ਹੈ।
f(60)=50। 60 ਮਿੰਟਾਂ ਵਿੱਚ, 50 ਮੀਲ ਦੀ ਯਾਤਰਾ ਕੀਤੀ ਜਾਂਦੀ ਹੈ।
f −1 (60)=70। 60 ਮੀਲ ਦੀ ਯਾਤਰਾ ਕਰਨ ਲਈ, 70 ਮਿੰਟ ਲੱਗਣਗੇ।
a. 3; b. 5.6
x=3y+5
f −1 (x)= ( 2−x ) 2 ; f ਦਾ ਡੋਮੇਨ:[ 0,∞ ); f −1 ਦਾ ਡੋਮੇਨ:( −∞,2 ]
1.1 ਸੈਕਸ਼ਨ ਅਭਿਆਸ
ਸੰਬੰਧ ਕ੍ਰਮਬੱਧ ਜੋੜਿਆਂ ਦਾ ਇੱਕ ਸਮੂਹ ਹੈ। ਇੱਕ ਫੰਕਸ਼ਨ ਸੰਬੰਧ ਦਾ ਇੱਕ ਵਿਸ਼ੇਸ਼ ਰੂਪ ਹੈ ਜਿਸ ਵਿੱਚ ਕੋਈ ਵੀ ਦੋ ਕ੍ਰਮਬੱਧ ਜੋੜੇ ਇੱਕੋ ਪਹਿਲੇ ਕੋਆਰਡੀਨੇਟ ਨਹੀਂ ਰੱਖਦੇ।
ਜਦੋਂ ਇੱਕ ਲੰਬਕਾਰੀ ਰੇਖਾ ਕਿਸੇ ਸੰਬੰਧ ਦੇ ਗ੍ਰਾਫ਼ ਨੂੰ ਇੱਕ ਤੋਂ ਵੱਧ ਵਾਰ ਕੱਟਦੀ ਹੈ, ਤਾਂ ਇਹ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਉਸ ਇਨਪੁਟ ਲਈ ਇੱਕ ਤੋਂ ਵੱਧ ਆਉਟਪੁਟ ਹਨ। ਕਿਸੇ ਵੀ ਖਾਸ ਇਨਪੁਟ ਮੁੱਲ 'ਤੇ, ਸਿਰਫ਼ ਇੱਕ ਆਉਟਪੁਟ ਹੋ ਸਕਦਾ ਹੈ ਜੇਕਰ ਸੰਬੰਧ ਨੂੰ ਫੰਕਸ਼ਨ ਬਣਨਾ ਹੈ।
ਜਦੋਂ ਇੱਕ ਖਿਤਿਜੀ ਰੇਖਾ ਕਿਸੇ ਫੰਕਸ਼ਨ ਦੇ ਗ੍ਰਾਫ਼ ਨੂੰ ਇੱਕ ਤੋਂ ਵੱਧ ਵਾਰ ਕੱਟਦੀ ਹੈ, ਤਾਂ ਇਹ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਉਸ ਆਉਟਪੁਟ ਲਈ ਇੱਕ ਤੋਂ ਵੱਧ ਇਨਪੁਟ ਹਨ। ਇੱਕ ਫੰਕਸ਼ਨ ਇੱਕ-ਤੋਂ-ਇੱਕ ਹੁੰਦਾ ਹੈ ਜੇਕਰ ਹਰੇਕ ਆਉਟਪੁਟ ਸਿਰਫ਼ ਇੱਕ ਇਨਪੁਟ ਨਾਲ ਮੇਲ ਖਾਂਦਾ ਹੈ।
ਫੰਕਸ਼ਨ
ਫੰਕਸ਼ਨ
function
function
function
function
function
function
function
not a function
f(−3)=−11; f(−3)=−11; f(2)=−1; f(2)=−1; f(−a)=−2a−5; f(−a)=−2a−5; −f(a)=−2a+5;−f(a)=−2a+5; f(a+h)=2a+2h−5 f(a+h)=2a+2h−5
f(−3)= 5 +5; f(−3)= 5 +5; f(2)=5; f(2)=5; f(−a)= 2+a +5; f(−a)= 2+a +5; −f(a)=− 2−a −5;−f(a)=− 2−a −5;f(a+h)= 2−a−h +5 f(a+h)= 2−a−h +5
f(−3)=2;f(−3)=2; f(2)=1−3=−2; f(2)=1−3=−2; f(−a)=| −a−1 |−| −a+1 |; f(−a)=|−a−1|−|−a+1|; −f(a)=−| a−1 |+| a+1 |;−f(a)=−|a−1|+|a+1|; f(a+h)=| a+h−1 |−| a+h+1 | f(a+h)=|a+h−1|−|a+h+1|
g(x)−g(a) x−a =x+a+2,x≠a g(x)−g(a) x−a =x+a+2,x≠a
ⓐ f(−2)=14; f(−2)=14;
ⓑ x=3 x=3
ⓐ f(5)=10; f(5)=10;
ⓑ x=−1 x=−1 or x=4 x=4
ⓐ f(t)=6− 2 3 t; f(t)=6− 2 3 t;
ⓑ f(−3)=8; f(−3)=8;
ⓒ t=6 t=6
not a function
function
function
function
function
ਕਾਰਜ
f(0)=1; f(0)=1;
f(x)=−3,x=−2 ਜਾਂ x=2
ਕਾਰਜ ਨਹੀਂ ਹੈ, ਇਸ ਲਈ ਇਹ ਇੱਕ-ਇੱਕ-ਵਾਲਾ ਕਾਰਜ ਵੀ ਨਹੀਂ ਹੈ
ਇੱਕ-ਇੱਕ-ਵਾਲਾ ਕਾਰਜ
ਕਾਰਜ, ਪਰ ਇੱਕ-ਇੱਕ-ਵਾਲਾ ਨਹੀਂ
ਕਾਰਜ
ਕਾਰਜ
ਕਾਰਜ ਨਹੀਂ
f(x)=1,x=2
f(−2)=14; f(−1)=11; f(0)=8; f(1)=5; f(2)=2
f(−2)=4; f(−1)=4.414; f(0)=4.732; f(1)=5; f(2)=5.236
f(−2)= 1/9 ; f(−1)= 1/3 ; f(0)=1; f(1)=3; f(2)=9
20
[0, 100]
[−0.001, 0.001]
[−1,000,000, 1,000,000]
[0, 10]
[−0.1,0.1]
[−100, 100]
g(5000)=50; g(5000)=50;
100 ਵਰਗ ਫੁੱਟ ਦੇ ਬਾਗ ਲਈ ਲੋੜੀਂਦੇ ਮਿੱਟੀ ਦੇ ਘਣ ਗਜ਼ਾਂ ਦੀ ਗਿਣਤੀ 1 ਹੈ।
1 ਸੈਕਿੰਡ ਬਾਅਦ ਜ਼ਮੀਨ ਤੋਂ ਉੱਪਰ ਇੱਕ ਰਾਕੇਟ ਦੀ ਉਚਾਈ 200 ਫੁੱਟ ਹੈ।
2 ਸੈਕਿੰਡ ਬਾਅਦ ਜ਼ਮੀਨ ਤੋਂ ਉੱਪਰ ਇੱਕ ਰਾਕੇਟ ਦੀ ਉਚਾਈ 350 ਫੁੱਟ ਹੈ।
1. 1.2 ਭਾਗ ਅਭਿਆਸ
2. ਕਿਸੇ ਫਲਨ (function) ਦਾ ਪ੍ਰਦੇਸ਼ (domain) ਉਨ੍ਹਾਂ ਸੁਤੰਤਰ ਪਰਿਮਾਣਾਂ (independent variables) ਦੇ ਮੁੱਲਾਂ 'ਤੇ ਨਿਰਭਰ ਕਰਦਾ ਹੈ ਜੋ ਫਲਨ ਨੂੰ ਅਨਿਸ਼ਚਿਤ (undefined) ਜਾਂ ਕਾਲਪਨਿਕ (imaginary) ਬਣਾਉਂਦੇ ਹਨ।
3. f(x)= x 3 f(x)= x 3 ਲਈ x x 'ਤੇ ਕੋਈ ਪਾਬੰਦੀ ਨਹੀਂ ਹੈ ਕਿਉਂਕਿ ਤੁਸੀਂ ਕਿਸੇ ਵੀ ਵਾਸਤਵਿਕ ਸੰਖਿਆ (real number) ਦਾ ਘਣਮੂਲ (cube root) ਲੈ ਸਕਦੇ ਹੋ। ਇਸ ਲਈ ਪ੍ਰਦੇਸ਼ ਸਾਰੀਆਂ ਵਾਸਤਵਿਕ ਸੰਖਿਆਵਾਂ, (−∞,∞) (−∞,∞) ਹੈ। ਵਾਸਤਵਿਕ ਸੰਖਿਆਵਾਂ ਦੇ ਸਮੂਹ ਨਾਲ ਨਜਿੱਠਣ ਵੇਲੇ, ਤੁਸੀਂ ਨਕਾਰਾਤਮਕ ਸੰਖਿਆਵਾਂ ਦਾ ਵਰਗਮੂਲ (square root) ਨਹੀਂ ਲੈ ਸਕਦੇ। ਇਸ ਲਈ f(x)= x f(x)= x ਲਈ x x-ਮੁੱਲਾਂ ਨੂੰ ਅਣ-ਨਕਾਰਾਤਮਕ (nonnegative) ਸੰਖਿਆਵਾਂ ਤੱਕ ਸੀਮਤ ਕੀਤਾ ਗਿਆ ਹੈ ਅਤੇ ਪ੍ਰਦੇਸ਼ [0,∞) [0,∞) ਹੈ।
4. ਟੁਕੜਿਆਂ ਵਾਲੇ ਫਲਨ (piecewise function) ਦੇ ਹਰੇਕ ਸੂਤਰ ਨੂੰ ਉਸਦੇ ਸੰਬੰਧਿਤ ਪ੍ਰਦੇਸ਼ ਉੱਤੇ ਗ੍ਰਾਫ ਕਰੋ। ਹਰੇਕ ਗ੍ਰਾਫ ਲਈ x x-ਧੁਰੇ ਅਤੇ y y-ਧੁਰੇ ਲਈ ਇੱਕੋ ਪੈਮਾਨਾ (scale) ਵਰਤੋ। ਸੰਮਿਲਤ ਅੰਤਿਮ ਬਿੰਦੂਆਂ (inclusive endpoints) ਨੂੰ ਇੱਕ ਸੰਪੂਰਨ ਚੱਕਰ (solid circle) ਨਾਲ ਅਤੇ ਅਸੰਮਿਲਤ ਅੰਤਿਮ ਬਿੰਦੂਆਂ (exclusive endpoints) ਨੂੰ ਇੱਕ ਖਾਲੀ ਚੱਕਰ (open circle) ਨਾਲ ਦਰਸਾਓ। −∞ −∞ ਜਾਂ ∞ ∞ ਦਰਸਾਉਣ ਲਈ ਇੱਕ ਤੀਰ (arrow) ਦੀ ਵਰਤੋਂ ਕਰੋ। ਟੁਕੜਿਆਂ ਵਾਲੇ ਫਲਨ ਦੇ ਗ੍ਰਾਫ ਨੂੰ ਲੱਭਣ ਲਈ ਗ੍ਰਾਫਾਂ ਨੂੰ ਜੋੜੋ।
5. (−∞,∞) (−∞,∞)
6. (−∞,3] (−∞,3]
7. (−∞,∞) (−∞,∞)
8. (−∞,∞) (−∞,∞)
9. (−∞,− 1 2 )∪(− 1 2 ,∞) (−∞,− 1 2 )∪(− 1 2 ,∞)
10. (−∞,−11)∪(−11,2)∪(2,∞) (−∞,−11)∪(−11,2)∪(2,∞)
11. (−∞,−3)∪(−3,5)∪(5,∞) (−∞,−3)∪(−3,5)∪(5,∞)
12. (−∞,5) (−∞,5)
13. [6,∞) [6,∞)
14. ( −∞,−9 )∪( −9,9 )∪( 9,∞ ) ( −∞,−9 )∪( −9,9 )∪( 9,∞ )
15. ਪ੍ਰਦੇਸ਼: (2,8], (2,8], ਪਰਿਸਰ (range) [6,8) [6,8)
16. ਪ੍ਰਦੇਸ਼: [−4, 4], [−4, 4], ਪਰਿਸਰ: [0, 2] [0, 2]
17. ਪ੍ਰਦੇਸ਼: [−5,3), [−5,3), ਪਰਿਸਰ: [ 0,2 ] [ 0,2 ]
18. ਪ੍ਰਦੇਸ਼: (−∞,1], (−∞,1], ਪਰਿਸਰ: [0,∞) [0,∞)
19. ਪ੍ਰਦੇਸ਼: [ −6,− 1 6 ]∪[ 1 6 ,6 ]; [ −6,− 1 6 ]∪[ 1 6 ,6 ]; ਪਰਿਸਰ: [ −6,− 1 6 ]∪[ 1 6 ,6 ] [ −6,− 1 6 ]∪[ 1 6 ,6 ]
20. ਪ੍ਰਦੇਸ਼: [−3,∞); [−3,∞); ਪਰਿਸਰ: [0,∞) [0,∞)
21. ਪ੍ਰਦੇਸ਼: (−∞,∞) (−∞,∞)
22. ਪ੍ਰਦੇਸ਼: (−∞,∞) (−∞,∞)
23. ਪ੍ਰਦੇਸ਼: (−∞,∞) (−∞,∞)
24. ਪ੍ਰਦੇਸ਼: (−∞,∞) (−∞,∞)
f(−3)=1; f(−2)=0; f(−1)=0; f(0)=0
f(−1)=−4; f(0)=6; f(2)=20; f(4)=34
f(−1)=−5; f(0)=3; f(2)=3; f(4)=16
ਡੋਮੇਨ: (−∞,1)∪(1,∞)
ਵਿੰਡੋ: [−0.5,−0.1]; ਰੇਂਜ: [4,100]
ਵਿੰਡੋ: [0.1,0.5]; ਰੇਂਜ: [4,100]
[0,8]
ਬਹੁਤ ਸਾਰੇ ਜਵਾਬ। ਇੱਕ ਫੰਕਸ਼ਨ ਹੈ f(x)= 1 x−2 ।
1.3 ਭਾਗ ਅਭਿਆਸ
ਹਾਂ, ਸਾਰੇ ਰੇਖੀ ਫੰਕਸ਼ਨਾਂ ਦੀ ਔਸਤ ਤਬਦੀਲੀ ਦਰ ਸਥਿਰ ਹੁੰਦੀ ਹੈ।
ਸੰਪੂਰਨ ਵੱਧ ਤੋਂ ਵੱਧ ਅਤੇ ਘੱਟ ਤੋਂ ਘੱਟ ਸਾਰੇ ਗ੍ਰਾਫ ਨਾਲ ਸਬੰਧਤ ਹਨ, ਜਦੋਂ ਕਿ ਸਥਾਨਕ ਐਕਸਟ੍ਰੀਮਾ ਸਿਰਫ ਇੱਕ ਖੁੱਲ੍ਹੇ ਅੰਤਰਾਲ ਦੇ ਆਸਪਾਸ ਇੱਕ ਖਾਸ ਖੇਤਰ ਨਾਲ ਸਬੰਧਤ ਹਨ।
4( b+1 )
3
4x+2h
−1 13( 13+h )
3 h 2 +9h+9
4x+2h−3
4 3
( −∞,−2.5 )∪( 1,∞ ) 'ਤੇ ਵੱਧ ਰਿਹਾ ਹੈ, (−2.5,1) 'ਤੇ ਘੱਟ ਰਿਹਾ ਹੈ।
( −∞,1 )∪( 3,4 ) 'ਤੇ ਵੱਧ ਰਿਹਾ ਹੈ, ( 1,3 )∪( 4,∞ ) 'ਤੇ ਘੱਟ ਰਿਹਾ ਹੈ।
ਸਥਾਨਕ ਵੱਧ ਤੋਂ ਵੱਧ: (−3,50), ਸਥਾਨਕ ਘੱਟ ਤੋਂ ਘੱਟ: (3,−50)
ਲਗਭਗ (7,150) 'ਤੇ ਸੰਪੂਰਨ ਵੱਧ ਤੋਂ ਵੱਧ, ਲਗਭਗ (−7.5,−220) 'ਤੇ ਸੰਪੂਰਨ ਘੱਟ ਤੋਂ ਘੱਟ।
a. –3000; b. –1250
-4
27
–0.167
Local minimum at (3,−22), (3,−22), decreasing on (−∞,3), (−∞,3), increasing on (3,∞) (3,∞)
Local minimum at (−2,−2), (−2,−2), decreasing on (−3,−2), (−3,−2), increasing on (−2,∞) (−2,∞)
Local maximum at (−0.39,5.98), (−0.39,5.98), local minima at (−3.15,−47.62) (−3.15,−47.62) and (2.04,-32.04), (2.04,-32.04), decreasing on (−∞,−3.15)∪ (−0.39,2.04), (−∞,−3.15)∪ (−0.39,2.04), increasing on (−3.15,−0.39)∪ (2.04,∞) (−3.15,−0.39)∪ (2.04,∞)
A
b=5 b=5
2.7 gallons per minute
approximately –0.6 milligrams per day
1.4 Section Exercises
Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g, g, such as an even-indexed root or zeros in the denominator.
Yes. Sample answer: Let f(x)=x+1 and g(x)=x−1. f(x)=x+1 and g(x)=x−1. Then f(g(x))=f(x−1)=(x−1)+1=x f(g(x))=f(x−1)=(x−1)+1=x and g(f(x))=g(x+1)=(x+1)−1=x. g(f(x))=g(x+1)=(x+1)−1=x. So f∘g=g∘f. f∘g=g∘f.
(f+g)( x )=2x+6, (f+g)( x )=2x+6, domain: (−∞,∞) (−∞,∞)
(f−g)( x )=2 x 2 +2x−6, (f−g)( x )=2 x 2 +2x−6, domain: (−∞,∞) (−∞,∞)
(fg)( x )=− x 4 −2 x 3 +6 x 2 +12x, (fg)( x )=− x 4 −2 x 3 +6 x 2 +12x, domain: (−∞,∞) (−∞,∞)
( f g )( x )= x 2 +2x 6− x 2 , ( f g )( x )= x 2 +2x 6− x 2 , domain: (−∞,− 6 )∪(− 6 , 6 )∪( 6 ,∞) (−∞,− 6 )∪(− 6 , 6 )∪( 6 ,∞)
(f+g)( x )= 4 x 3 +8 x 2 +1 2x , (f+g)( x )= 4 x 3 +8 x 2 +1 2x , domain: (−∞,0)∪(0,∞) (−∞,0)∪(0,∞)
(f−g)( x )= 4 x 3 +8 x 2 −1 2x , (f−g)( x )= 4 x 3 +8 x 2 −1 2x , domain: (−∞,0)∪(0,∞) (−∞,0)∪(0,∞)
(fg)( x )=x+2, (fg)( x )=x+2, domain: (−∞,0)∪(0,∞) (−∞,0)∪(0,∞)
( f g )( x )=4 x 3 +8 x 2 , ( f g )( x )=4 x 3 +8 x 2 , domain: (−∞,0)∪(0,∞) (−∞,0)∪(0,∞)
(f+g)(x)=3 x 2 + x−5 , (f+g)(x)=3 x 2 + x−5 , domain: [5,∞) [5,∞)
(f−g)(x)=3 x 2 − x−5 , (f−g)(x)=3 x 2 − x−5 , domain: [5,∞) [5,∞)
(fg)(x)=3 x 2 x−5 , (fg)(x)=3 x 2 x−5 , domain: [5,∞) [5,∞)
( f g )(x)= 3 x 2 x−5 , ( f g )(x)= 3 x 2 x−5 , domain: (5,∞) (5,∞)
ⓐ 3
ⓑ f( g( x ) )=2 ( 3x−5 ) 2 +1; f( g( x ) )=2 ( 3x−5 ) 2 +1;
ⓒ g( f)( x ) )=6 x 2 −2; g( f)( x ) )=6 x 2 −2;
ⓓ ( g∘g )(x)=3(3x−5)−5=9x−20; ( g∘g )(x)=3(3x−5)−5=9x−20;
ⓔ ( f∘f )( −2 )=163 ( f∘f )( −2 )=163
f(g(x))= x 2 +3 +2,g(f(x))=x+4 x +7 f(g(x))= x 2 +3 +2,g(f(x))=x+4 x +7
f(g(x))= x+1 x 3 3 = x+1 3 x ,g(f(x))= x 3 +1 x f(g(x))= x+1 x 3 3 = x+1 3 x ,g(f(x))= x 3 +1 x
( f∘g )(x)= 1 2 x +4−4 = x 2 ,( g∘f )(x)=2x−4 ( f∘g )(x)= 1 2 x +4−4 = x 2 ,( g∘f )(x)=2x−4
f(g(h(x)))= ( 1 x+3 ) 2 +1 f(g(h(x)))= ( 1 x+3 ) 2 +1
ⓐ Text (g∘f)(x)=− 3 2−4x ; (g∘f)(x)=− 3 2−4x ;
ⓑ( −∞, 1 2 ) ( −∞, 1 2 )
ⓐ (0,2)∪(2,∞); (0,2)∪(2,∞);
ⓑ (−∞,−2)∪(2,∞); (−∞,−2)∪(2,∞); c. (0,∞) (0,∞)
(1,∞) (1,∞)
sample: f(x)= x 3 g(x)=x−5 f(x)= x 3 g(x)=x−5
sample: f(x)= 4 x g(x)= (x+2) 2 f(x)= 4 x g(x)= (x+2) 2
sample: f(x)= x 3 g(x)= 1 2x−3 f(x)= x 3 g(x)= 1 2x−3
sample: f(x)= x 4 g(x)= 3x−2 x+5 f(x)= x 4 g(x)= 3x−2 x+5
sample: f(x)= x f(x)= x g(x)=2x+6 g(x)=2x+6
sample: f(x)= x 3 f(x)= x 3 g(x)=(x−1) g(x)=(x−1)
sample: f(x)= x 3 f(x)= x 3 g(x)= 1 x−2 g(x)= 1 x−2
sample: f(x)= x f(x)= x g(x)= 2x−1 3x+4 g(x)= 2x−1 3x+4
2
5
4
0
2
1
4
4
9
4
2
3
11
0
7
f(g(0))=27,g( f(0) )=−94 f(g(0))=27,g( f(0) )=−94
f(g(0))= 1 5 ,g(f(0))=5 f(g(0))= 1 5 ,g(f(0))=5
18 x 2 +60x+51 18 x 2 +60x+51
g∘g(x)=9x+20 g∘g(x)=9x+20
2
(−∞,∞) (−∞,∞)
False
(f∘g)(6)=6 (f∘g)(6)=6; (g∘f)(6)=6 (g∘f)(6)=6
(f∘g)(11)=11,(g∘f)(11)=11 (f∘g)(11)=11,(g∘f)(11)=11
c
A(t)=π ( 25 t+2 ) 2 A(t)=π ( 25 t+2 ) 2 and A(2)=π ( 25 4 ) 2 =2500π A(2)=π ( 25 4 ) 2 =2500π square inches
A(5)=π ( 2(5)+1 ) 2 =121π A(5)=π ( 2(5)+1 ) 2 =121π square units
ⓐ N(T(t))=23 (5t+1.5) 2 −56(5t+1.5)+1; N(T(t))=23 (5t+1.5) 2 −56(5t+1.5)+1;
ⓑ 3.38 hours
1.5 Section Exercises
A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.
A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.
For a function f, f, substitute (−x) (−x) for (x) (x) in f(x). f(x). Simplify. If the resulting function is the same as the original function, f(−x)=f(x), f(−x)=f(x), then the function is even. If the resulting function is the opposite of the original function, f(−x)=−f(x), f(−x)=−f(x), then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.
g(x)=|x-1|−3 g(x)=|x-1|−3
g(x)= 1 (x+4) 2 +2 g(x)= 1 (x+4) 2 +2
The graph of f(x+43) f(x+43) is a horizontal shift to the left 43 units of the graph of f. f.
The graph of f(x-4) f(x-4) is a horizontal shift to the right 4 units of the graph of f. f.
The graph of f(x)+8 f(x)+8 is a vertical shift up 8 units of the graph of f. f.
The graph of f(x)−7 f(x)−7 is a vertical shift down 7 units of the graph of f. f.
The graph of f(x+4)−1 f(x+4)−1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f. f.
decreasing on (−∞,−3) (−∞,−3) and increasing on (−3,∞) (−3,∞)
decreasing on [0,∞) [0,∞)
g(x)=f(x-1),h(x)=f(x)+1 g(x)=f(x-1),h(x)=f(x)+1
f(x)=|x-3|−2 f(x)=|x-3|−2
f(x)= x+3 −1 f(x)= x+3 −1
f(x)= (x-2) 2 f(x)= (x-2) 2
f(x)=|x+3|−2 f(x)=|x+3|−2
f(x)=− x f(x)=− x
f(x)=− (x+1) 2 +2 f(x)=− (x+1) 2 +2
f(x)= −x +1 f(x)= −x +1
even
odd
even
The graph of g g is a vertical reflection (across the x x-axis) of the graph of f. f.
The graph of g g is a vertical stretch by a factor of 4 of the graph of f. f.
The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f. f.
The graph of g g is a horizontal stretch by a factor of 3 of the graph of f. f.
The graph of g g is a horizontal reflection across the y y-axis and a vertical stretch by a factor of 3 of the graph of f. f.
g(x)=|−4x| g(x)=|−4x|
g(x)= 1 3 (x+2) 2 −3 g(x)= 1 3 (x+2) 2 −3
g(x)= 1 2 (x-5) 2 +1 g(x)= 1 2 (x-5) 2 +1
The graph of the function f(x)= x 2 f(x)= x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.
The graph of f(x)=|x| f(x)=|x| is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.
The graph of the function f(x)= x 3 f(x)= x 3 is compressed vertically by a factor of 1 2 . 1 2 .
The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.
The graph of f(x)= x f(x)= x is reflected across the y-axis and then shifted right 4 units.
1.6 Section Exercises
Isolate the absolute value term so that the equation is of the form |A|=B. |A|=B. Form one equation by setting the expression inside the absolute value symbol, A, A, equal to the expression on the other side of the equation, B. B. Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, −B. −B. Solve each equation for the variable.
The graph of the absolute value function does not cross the x x-axis, so the graph is either completely above or completely below the x x-axis.
First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.
| x+4 |= 1 2 | x+4 |= 1 2
|f(x)−8|<0.03 |f(x)−8|<0.03
{ 1,11 } { 1,11 }
{ - 9 4 , 13 4 } { - 9 4 , 13 4 }
{ 10 3 , 20 3 } { 10 3 , 20 3 }
{ 11 5 , 29 5 } { 11 5 , 29 5 }
{ 5 2 , 7 2 } { 5 2 , 7 2 }
No solution
{ −57,27 } { −57,27 }
( 0,−8 );( −6,0 ),( 4,0 ) ( 0,−8 );( −6,0 ),( 4,0 )
( 0,−7 ); ( 0,−7 ); no x x-intercepts
(−∞,−8)∪(12,∞) (−∞,−8)∪(12,∞)
−43,4−43,4
( −∞,− 8 3 ]∪[ 6,∞ ) ( −∞,− 8 3 ]∪[ 6,∞ )
( −∞,− 8 3 ]∪[ 16,∞ ) ( −∞,− 8 3 ]∪[ 16,∞ )
range: [ 0,20 ] [ 0,20 ]
x- x- intercepts:
(−∞,∞) (−∞,∞)
There is no solution for a a that will keep the function from having a y y-intercept. The absolute value function always crosses the y y-intercept when x=0. x=0.
| p−0.08 |≤0.015 | p−0.08 |≤0.015
| x−5.0 |≤0.01 | x−5.0 |≤0.01
1.7 Section Exercises
Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y-values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y-values repeat and the function is one-to-one.
Yes. For example, f(x)= 1 x f(x)= 1 x is its own inverse.
Given a function y=f(x), y=f(x), solve for x x in terms of y. y. Interchange the x x and y. y. Solve the new equation for y. y. The expression for y y is the inverse, y= f −1 (x). y= f −1 (x).
f −1 (x)=x−3 f −1 (x)=x−3
f −1 (x)=2−x f −1 (x)=2−x
f −1 (x)= −2x x−1 f −1 (x)= −2x x−1
domain of f(x):[−7,∞); f −1 (x)= x −7 f(x):[−7,∞); f −1 (x)= x −7
domain of f(x):[0,∞); f −1 (x)= x+5 f(x):[0,∞); f −1 (x)= x+5
ⓐ f(g(x))=x f(g(x))=x and g(f(x))=x. g(f(x))=x.
ⓑ This tells us that f f and g g are inverse functions
f(g(x))=x,g(f(x))=x f(g(x))=x,g(f(x))=x
one-to-one
one-to-one
not one-to-one
3 3
2 2
[ 2,10 ] [ 2,10 ]
6 6
−4 −4
0 0
1 1
row: x x | 1 | 4 | 7 | 12 | 16
row: f −1 (x) f −1 (x) | 3 | 6 | 9 | 13 | 14
f −1 (x)= (1+x) 1/3 f −1 (x)= (1+x) 1/3
f −1 (x)= 5 9 ( x−32 ). f −1 (x)= 5 9 ( x−32 ). Given the Fahrenheit temperature, x, x, this formula allows you to calculate the Celsius temperature.
t(d)= d 50 , t(d)= d 50 , t(180)= 180 50 . t(180)= 180 50 . The time for the car to travel 180 miles is 3.6 hours.
Review Exercises
function
not a function
f(−3)=−27; f(−3)=−27; f(2)=−2; f(2)=−2; f(−a)=−2 a 2 −3a; f(−a)=−2 a 2 −3a; −f(a)=2 a 2 −3a; −f(a)=2 a 2 −3a; f(a+h)=−2 a 2 +3a−4ah+3h−2 h 2 f(a+h)=−2 a 2 +3a−4ah+3h−2 h 2
one-to-one
function
function
2 2
x=−1.8 x=−1.8 or or x=1.8 or x=1.8
−64+80a−16 a 2 −1+a =−16a+64 −64+80a−16 a 2 −1+a =−16a+64
( −∞,−2 )∪( −2,6 )∪( 6,∞ ) ( −∞,−2 )∪( −2,6 )∪( 6,∞ )
31 31
increasing ( 2,∞ ); ( 2,∞ ); decreasing (−∞,2) (−∞,2)
increasing ( −3,1 ); ( −3,1 ); constant (−∞,−3)∪( 1,∞ ) (−∞,−3)∪( 1,∞ )
local minimum ( −2,−3 ); ( −2,−3 ); local maximum ( 1,3 ) ( 1,3 )
Absolute Maximum: 10
( f∘g )(x)=17−18x;( g∘f )(x)=−7−18x ( f∘g )(x)=17−18x;( g∘f )(x)=−7−18x
( f∘g )(x)= 1 x +2 ; ( f∘g )(x)= 1 x +2 ;( g∘f )(x)= 1 x+2 ( g∘f )(x)= 1 x+2
(f∘g)(x)= 1+x 1+4x ,x≠0,x≠− 1 4 (f∘g)(x)= 1+x 1+4x ,x≠0,x≠− 1 4
( f∘g )(x)= 1 x ,x>0 ( f∘g )(x)= 1 x ,x>0
sample: g(x)= 2x−1 3x+4 ;f(x)= x g(x)= 2x−1 3x+4 ;f(x)= x
f(x)=| x−3 | f(x)=| x−3 |
even
odd
even
f(x)= 1 2 | x+2 |+1 f(x)= 1 2 | x+2 |+1
f(x)=−3| x−3 |+3 f(x)=−3| x−3 |+3
x=−22,x=14 x=−22,x=14
( − 5 3 ,3 ) ( − 5 3 ,3 )
f −1 (x) = x-9 10 f −1 (x) = x-9 10
The function is one-to-one.
The function is not one-to-one.
5 5
Practice Test
The relation is a function.
−16
The graph is a parabola and the graph fails the horizontal line test.
2 a 2 −a 2 a 2 −a
−2(a+b)+1 −2(a+b)+1
2 2
even even
odd odd
x=−7 x=−7 and x=10 x=10
f −1 (x)= x+5 3 f −1 (x)= x+5 3
(−∞,−1.1) and (1.1,∞) (−∞,−1.1) and (1.1,∞)
( 1.1,−0.9 ) ( 1.1,−0.9 )
f(2)=2 f(2)=2
f(x)={ | x |ifx≤2 3ifx>2 f(x)={ | x |ifx≤2 3ifx>2
x=2 x=2
yes
f −1 (x)=− x−11 2 f −1 (x)=− x−11 2