Describe the monotony of $f:\mathbb{R} \rightarrow \mathbb{R}, \: f(x)=2^x3^{1-x}+2^{1-x}3^x.$ With derivatives this is a piece of cake, but I wonder what's the elementar method for solving it.
Monotony without use of derivatives
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exponential-function
monotone-functions
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0your function is not monoton – 2017-01-30
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0Hint: The function is a sum of 2 values whose product is constant. – 2017-01-30
1 Answers
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By a translation, the problem can be reduced to proving that $$x\in[0,\infty)\mapsto a^x + a^{-x}$$ is increasing for $a>1$. Let be $0\le x < y$. As $x\mapsto a^x$ is increasing, $a^{-x}-a^{-y} > 0$ and: $$a^{x+y} > 1\implies a^y - a^x = a^{x+y}(a^{-x} - a^{-y}) > a^{-x} - a^{-y}\implies a^y + a^{-y} > a^x + a^{-x}.$$