f: R—>R,f(x)= x²+4x+5, Is the given function one-one or onto?
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f : R -----> R , f(x) = x² + 4x + 5
To check function is one - one or many one , Let us consider two points x1 and x2 on R such that , f(x1) = f(x2)
f(x1) = x1² + 4x1 + 5
f(x2) = x2² + 4x2 + 5
now, f(x1) = f(x2)
x1² + 4x1 + 5 = x2² + 4x2 + 5
x1² + 4x1 = x2² + 4x2
(x1 - x2)(x1 + x2) + 4(x1 - x2) = 0
x1 + x2 + 4 = 0
here, x1 ≠ x2
hence, it is clear that function is not one one.
To check function is onto or not , we have to find range of function.
y = x² + 4x + 5
x² + 4x + 5 -y = 0
D = 4² - 4(5 - y) ≥ 0 for all real value of x
= 4 - 5 + y ≥ 0
= -1 + y ≥ 0
= y ≥ 1 , hence, range
(1, ∞)
here, co-domain ≠ range
so, function is not onto.
e.g., f(x) is neither one one nor onto.
To check function is one - one or many one , Let us consider two points x1 and x2 on R such that , f(x1) = f(x2)
f(x1) = x1² + 4x1 + 5
f(x2) = x2² + 4x2 + 5
now, f(x1) = f(x2)
x1² + 4x1 + 5 = x2² + 4x2 + 5
x1² + 4x1 = x2² + 4x2
(x1 - x2)(x1 + x2) + 4(x1 - x2) = 0
x1 + x2 + 4 = 0
here, x1 ≠ x2
hence, it is clear that function is not one one.
To check function is onto or not , we have to find range of function.
y = x² + 4x + 5
x² + 4x + 5 -y = 0
D = 4² - 4(5 - y) ≥ 0 for all real value of x
= 4 - 5 + y ≥ 0
= -1 + y ≥ 0
= y ≥ 1 , hence, range
here, co-domain ≠ range
so, function is not onto.
e.g., f(x) is neither one one nor onto.
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Answer:
The given function is neither one-one nor onto.
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