The Range of the function f(x)=ln (3x^2 +4) is :

Answers (1)

Solution:    We shall use that  the result  R(f)=D(f^-1).

                              y=ln (3x^2+4)Rightarrow (3x^2+4)=e^y\ \Rightarrow hspace0.5cmx^2=frace^y-43geq 0Rightarrow e^y-4geq 0,\ \Rightarrowhspace0.5cm e^ygeq 4Rightarrow ygeq 2ln2.\ \	herefore R(f)=[2ln2,infty )

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