Page 31 - Algebra 1

Explanation:
1.
2.
Put x = a
f(a)= 𝑎 𝑎−2
Put x = a +2
f(a+2)= 𝑎+2 =𝑎+2 𝑎+2−2 𝑎
So,𝑓(𝑎) =𝑎 ×𝑎 =𝑎2 𝑓(𝑎+2) 𝑎−2 𝑎+2 𝑎2− 4
Lety=f(x)= 𝑥+1 2𝑥−3
y(2x – 3) = x + 1
2xy – 3y = x + 1
2xy – x = 3y + 1
x(2y – 1) = 3y + 1
x = 3𝑦+1 2y – 1
or f–1(x) = 3𝑥+1 2x – 1
Since x is real, x can take all values except for when 6x + 5 = 0
3.
−5 x=6
4. For a function to be even, f(– x) = f(x) f(x)=x2 +2x
f(x) =– f(x)
f(x)=|x|for(𝑥, 𝑥≥0).So,thefunctionisnotmodulus. –x, x<0
5. A function is when all 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.
f = {(0, 2), (1,3)}
6x = – 5
So, domain of the function is R - {−5} 6
f(–x)=(–x)2 +2(–x)=x2 – 2x
f(–x) ≠ f(x), so the function is not even
For a function to be odd,
–f(x)=–(x2 +2x)=–x2 –2x
f(x) ≠ –f(x), so the function is not odd
A function is constant when f(x) = c. So, this function is not constant.
A function is modulus when
Therefore, the above function is neither even, nor odd, nor modulus nor constant.
f = {(0, 2), (0,3), (1,4))}
0 of the first element doesn’t have a unique image in Q. So, this is not a function.
Page 30 of 54
ALGEBRA