Question

f: R—>R,f(x)= x²-x-2, Is the given function one-one or onto?

Answer 1

condition of one - one function :
if we take two different points x1 and x2 from domain of given function , f(x).
if we solve f(x1) = f(x2) , we get x1 = x2 then, f(x) is definitely an one one function.

let's take two different points x1 and x2
now, f(x1) = x1² - x1 - 2
f(x2) = x2² - x2 - 2

so, f(x1) = f(x2)
x1² - x1 - 2 = x2² - x2 - 2
x1² - x1 = x2² - x2
(x1 - x2)(x1 + x2) -(x1 - x2) = 0
(x1 - x2)(x1 + x2 - 1) = 0
x1 + x2 -1 = 0, (x1 - x2) = 0
x1 ≠ x2
hence, it is clear that function is not one - one.

condition of onto function :
co - domain = range

let's find range ,
y = x² - x - 2
x² - x - 2 - y = 0
discriminant = (-1)² - 4{-(2+y)} ≥ 0
1 + 4(2 + y) ≥ 0
1 + 8 + 4y ≥ 0
4y ≥ -9 => y ≥ -9/4
hence, range of function is [-9/4 , ∞)
hence, co-domain ≠ range
so, f(x) is not onto function.

hence, function,f(x) is neither one - one nor onto.

Answer 2

Answer:

Step-by-step explanation:

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