Question Number 31865 by rahul 19 last updated on 16/Mar/18
$${Range}\:{of}\:{function}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{6}^{{x}} +\mathrm{3}^{{x}} +\mathrm{6}^{−{x}} +\mathrm{3}^{−{x}} +\mathrm{2}. \\ $$
Commented by rahul 19 last updated on 16/Mar/18
$${my}\:{try}\:: \\ $$$${f}\left({x}\right)=\:\left(\mathrm{6}^{\frac{{x}}{\mathrm{2}}} +\mathrm{6}^{\frac{−{x}}{\mathrm{2}}} \right)^{\mathrm{2}} +\:\left(\mathrm{3}^{\frac{{x}}{\mathrm{2}}} +\mathrm{3}^{\frac{−{x}}{\mathrm{2}}} \right)^{\mathrm{2}} −\mathrm{2}\geqslant\mathrm{6}. \\ $$$${hence}\:{range}\:{is}\:\left[−\mathrm{6},\infty\right). \\ $$
Commented by rahul 19 last updated on 16/Mar/18
$${but}\:{ans}.\:{given}\:{is}\:\left[\mathrm{6},\infty\right).\: \\ $$
Commented by Tinkutara last updated on 16/Mar/18
Apply AM-GM.
Commented by rahul 19 last updated on 16/Mar/18
$${plz}\:{tell}\:{where}\:{am}\:{i}\:{wrong}\:? \\ $$
Commented by Tinkutara last updated on 16/Mar/18
$${Because}\:{you}\:{assumed}\:{minimum} \\ $$$${value}\:{of}\:\mathrm{6}^{\frac{{x}}{\mathrm{2}}} +\mathrm{6}^{\frac{−{x}}{\mathrm{2}}} \:{to}\:{be}\:\mathrm{0}\:{which}\:{is}\:{not}. \\ $$$$\mathrm{6}^{\frac{{x}}{\mathrm{2}}} +\mathrm{6}^{\frac{−{x}}{\mathrm{2}}} \:{has}\:{minimum}\:{value}\:\mathrm{2}. \\ $$
Commented by rahul 19 last updated on 16/Mar/18
$${ok}\:{thanks}\:!! \\ $$
Answered by mrW2 last updated on 16/Mar/18
$${f}\left({x}\right)=\left(\mathrm{6}^{{x}} +\mathrm{6}^{−{x}} \right)+\left(\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \right)+\mathrm{2} \\ $$$$\geqslant\mathrm{2}+\mathrm{2}+\mathrm{2}=\mathrm{6} \\ $$
Commented by rahul 19 last updated on 16/Mar/18
$${yes}\:{sir},\:{i}\:{have}\:{got}\:{the}\:{ans}. \\ $$$${thank}\:{u}\:{sir}! \\ $$