Page 32 - Algebra 1

P. 32

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All the elements of the first set are not associated with the element of the second. So, this is not
a function.
f=
{(0, 2), (2,3), (2,4)}
2 of the first element don’t have a unique image in Q. So, this is not a function.
f=
{(0, 2), (1,3), (2,4)}
All the elements of the first set are associated with the elements of the second set. And, an element of the first set has a unique image in the second set. So, this is a function.
6. Replacex=3x–1inf(x)=x2 –4x+5
f(x–1)=(x–1)2 –4(x–1)+5=x2 –2x+1–4x+4+5 f(x–1)=x2 –6x+10
As per the question,
f(x) = f(x – 1)
x2 –4x+5=x2 –6x+10
– 4x + 6x = 10 – 5
2x = 5
x=5 2
7. Replace x by f(x) f(f(x)) = 𝑓(𝑥)+1
Now put f(x) = 𝑥+1 in the above, 𝑥−1
𝑥+1+1
f(f(x)) = 𝑥−1 𝑥+1
f(f(x)) = f(f(x)) = x
𝑓(𝑥)−1
−1 𝑥+1+𝑥−1
𝑥−1
𝑥−1 = 2𝑥 𝑥+1−𝑥+1 2
𝑥−1
9. Inafunction(f–g)(x)=f(x)–g(x) = (x2 – 4) – (x + 2)
= (x – 2) (x +2) – (x + 2)
= (x + 2) (x – 2 – 1)
= (x + 2) (x – 3) =x2 –3x+2x–6 = x2 –x–6
8.
Since the function is a squared function, the range will always be positive.
Therefore, the range of the function is [0, ∞)
Page 31 of 54
ALGEBRA
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