Which of the function defined below is one-one?

Question

Which of the function defined below is one-one? (A) f: (0, ∞) → R , f (x) = x2 - 4x + 3 (B) f :[0, ∞)→ R , f (x) = x2 + 4x - 5 (C) f: R → R , f (x) =ex + (1/ex) (D) f: R → R, f (x) = ℓ n(x2 + x +1)

Tardigrade Admin · Accepted Answer

By definition only f (x) = x2 + 4x - 5 with domain [0, ∞) is one to one.

Q. Which of the function defined below is one-one?

$f : (0, \infty) \to R, f (x) = x^{2} - 4 x + 3$

$f : [0, \infty) \to R, f (x) = x^{2} + 4 x - 5$

$f : R \to R, f (x) = e^{x} + \frac{1}{e ^{x}}$

$f : R \to R, f (x) = ℓ n (x^{2} + x + 1)$

By definition only $f (x) = x^{2} + 4 x - 5$ with domain $[0, \infty)$ is one to one.

Q. Which of the function defined below is one-one?

f:(0,∞)→R,f(x)=x2−4x+3

f:[0,∞)→R,f(x)=x2+4x−5

f:R→R,f(x)=ex+ex1​

f:R→R,f(x)=ℓn(x2+x+1)

By definition only f(x)=x2+4x−5 with domain [0,∞) is one to one.

$f : (0, \infty) \to R, f (x) = x^{2} - 4 x + 3$

$f : [0, \infty) \to R, f (x) = x^{2} + 4 x - 5$

$f : R \to R, f (x) = e^{x} + \frac{1}{e ^{x}}$

$f : R \to R, f (x) = ℓ n (x^{2} + x + 1)$

By definition only $f (x) = x^{2} + 4 x - 5$ with domain $[0, \infty)$ is one to one.