Page 63 - Algebra

f=
{(0, 2), (2,3), (2,4)}
2 of the first elements 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. Replace x = 3x – 1 in f(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(𝑥)−1
−1 𝑥+1+𝑥−1
𝑥−1
𝑥−1 = 2𝑥 𝑥+1−𝑥+1 2
𝑥−1
f(f(x)) = x
8.
9. In a function (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
10. It’s given that f(x) = g(x) x + 5 = 5x2 – 1
5x2 – 1 – x – 5 = 0
5x2 –x–6=0
5x2 +5x–6x–6=0 5x( x + 1) – 1( x + 1) = 0 (5x – 1) (x + 1) = 0
x = – 1 or 1 5
f(f(x)) =
Since the function is a squared function, the range will always be positive.
Therefore, the range of the function is [0, ∞)
Page 62 of 177
Algebra I & II