Matematické Fórum

stuart clark · 21. 06. 2014 20:00

If $f(x) = 2^x-4^x+6^x-8^x-9^x+12^x\;,$ where $x\in \mathbb{R}.$ Then Range of $f(x)$ is

laszky · 08. 03. 2018 01:55

There holds $f(x)\geq -1$ with $f(1)=-1$ .

Since $f(x)+1 = \bigr[4^x-3^x-1\bigr]\bigr[3^x-2^x-1\bigr] = \bigr[3(4^{x-1}-3^{x-1})+(4^{x-1}-1)\bigr] \bigr[2(3^{x-1}-2^{x-1})+(3^{x-1}-1)\bigr]$

both square brackets are positive for $x>1$ and negative for $x<1$ . Thus, their product is positive for $x\neq 1$ and vanishes for $x=1$ .

Matematické Fórum

#1 21. 06. 2014 20:00

Range of f(x)

#2 08. 03. 2018 01:55

Re: Range of f(x)

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