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The solution set of the inequality \({3^{3x}} – {5^{3x}} + 3\left( {{3^x} – {5^x}} \right) > 0\) is
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A. \(\left( { – \infty ;0} \right)\). B. \(\left( { – \infty ;0} \right]\). C. \(\left( {0; + \infty } \right)\). D. \(\left[{0;+\infty}\right)\)[{0;+\infty}\right)\)
The answer
The given inequality is equivalent to \({3^{3x}} + {3.3^x} > {5^{3x}} + {3.5^x} \Leftrightarrow f\left( {{3^x}} \right) > f\left( {{5^x}} \right)\).
Considering the function \(f\left( t \right) = {t^3} + 3t\) on the interval \(\left( {0; + \infty } \right)\), we have: \(f’ \left( t \right) = 3{t^2} + 3 > 0,\forall t > 0\).
Derive a covariate function on the interval \(\left( {0; + \infty } \right)\).
So from \(f\left( {{3^x}} \right) > f\left( {{5^x}} \right)\) we have \({3^x} > {5^x } \Leftrightarrow {\left( {\frac{5}{3}} \right)^x}