- Problem Statement
- ##f : (-\infty, -1]\Rightarrow(0,e^5]## defined by ##f(x) = e^{x^3-3x+2}##. Check if the function is bijective.

- Relevant Equations
- Let ##h(x) = x^3 - 3x +2##

To check if it is injective :

##h'(x) = 3(x^2-1)##

##\implies h'(x) \geq 0## for ##x \in (-\infty, -1]##

Thus, ##f(x)## is increasing over the given domain and thus is one-one.

To check if it is surjective :

Range of ##f(x) = (0, e^4]## but co-domain is ##(0, e^5]## thus the function is into and thus cannot be bijective.

My Question :

1.How was the range found?

Edit: Domain of ##f## corrected.

##h'(x) = 3(x^2-1)##

##\implies h'(x) \geq 0## for ##x \in (-\infty, -1]##

Thus, ##f(x)## is increasing over the given domain and thus is one-one.

To check if it is surjective :

Range of ##f(x) = (0, e^4]## but co-domain is ##(0, e^5]## thus the function is into and thus cannot be bijective.

My Question :

1.How was the range found?

Edit: Domain of ##f## corrected.

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