Harder domain of a function

Algebra Level 3

What is the domain of the function below? f : y = 2 x − 3 x f:y = \sqrt {{2^x} - {3^x}}

D ( f ) = ⟨ − 2 3 ; 2 3 ⟩ D\left( f \right) = \left\langle { - \frac{2}{3};\frac{2}{3}} \right\rangle D ( f ) = ( 0 ; 2 3 ) D\left( f \right) = \left( {0;\left. {\frac{2}{3}} \right)} \right. D ( f ) = ( − ∞ ; 0 ) D\left( f \right) = \left( { - \infty ;0} \right) D ( f ) = ( − 2 3 ; 2 3 ) D\left( f \right) = \left( { - \frac{2}{3};\left. {\frac{2}{3}} \right)} \right. D ( f ) = ( − 2 3 ; 0 ) D\left( f \right) = \left( { - \frac{2}{3};\left. 0 \right)} \right. D ( f ) = ⟨ 0 ; ∞ ) D\left( f \right) = \left\langle {0;\left. \infty \right)} \right. D ( f ) = ( 0 ; ∞ ) D\left( f \right) = \left( {0;\left. \infty \right)} \right. D ( f ) = ( − ∞ ; 0 ⟩ D\left( f \right) = \left( {\left. { - \infty ;0} \right\rangle } \right.

Tomáš Hauser by Tomáš Hauser , 21, Czech Republic

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1 solution

Zico Quintina
Jul 2, 2018

We can use the fact that ln ⁡ \ln is a monotonically increasing function to find the domain of f ( x ) f(x) . We need

2 x ≥ 3 x ln ⁡ 2 x ≥ ln ⁡ 3 x [ Since ln is monotonically increasing ] x ln ⁡ 2 − x ln ⁡ 3 ≥ 0 x ( ln ⁡ 2 3 ) ≥ 0 x ≤ 0 [ Since ln ⁡ 2 3 is negative ] \begin{array}{rllll} 2^x &\ge \ \ 3^x \\ \\ \ln 2^x &\ge \ \ \ln 3^x & & & \small \text{[ Since ln is monotonically increasing ]} \\ \\ x \ln 2 - x \ln 3 &\ge \ \ 0 \\ \\ x \left( \ln \dfrac{2}{3} \right) &\ge \ \ 0 \\ \\ x &\le \ \ 0 & & & \small \text{[ Since } \ln \frac{2}{3} \text{ is negative ]} \end{array}

So the domain of f ( x ) f(x) is ( − ∞ , 0 ] \boxed{(-\infty, 0]}

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