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Question Number 195511 by Calculusboy last updated on 04/Aug/23

Answered by mr W last updated on 04/Aug/23

there are infinite values for k+1.    curve x^y +y^x =8 is symmetric about   y=x. say it passes the point (a,a),  then 2a^a =8 ⇒a=2.  straight line x+y=k^2  intersects the  curve x^y +y^x =8 if k^2 ≥a+a=4, i.e.  if k≤−2 or k≥2. so we have  k+1≤−1 or k+1≥3

$${there}\:{are}\:{infinite}\:{values}\:{for}\:{k}+\mathrm{1}. \\ $$$$ \\ $$$${curve}\:{x}^{{y}} +{y}^{{x}} =\mathrm{8}\:{is}\:{symmetric}\:{about}\: \\ $$$${y}={x}.\:{say}\:{it}\:{passes}\:{the}\:{point}\:\left({a},{a}\right), \\ $$$${then}\:\mathrm{2}{a}^{{a}} =\mathrm{8}\:\Rightarrow{a}=\mathrm{2}. \\ $$$${straight}\:{line}\:{x}+{y}={k}^{\mathrm{2}} \:{intersects}\:{the} \\ $$$${curve}\:{x}^{{y}} +{y}^{{x}} =\mathrm{8}\:{if}\:{k}^{\mathrm{2}} \geqslant{a}+{a}=\mathrm{4},\:{i}.{e}. \\ $$$${if}\:{k}\leqslant−\mathrm{2}\:{or}\:{k}\geqslant\mathrm{2}.\:{so}\:{we}\:{have} \\ $$$${k}+\mathrm{1}\leqslant−\mathrm{1}\:{or}\:{k}+\mathrm{1}\geqslant\mathrm{3} \\ $$

Commented by Calculusboy last updated on 04/Aug/23

thanks sir

$${thanks}\:{sir} \\ $$

Commented by kapoorshah last updated on 05/Aug/23

proof that x^y  + y^x  = 8 is symmetric  about y = x

$${proof}\:{that}\:{x}^{{y}} \:+\:{y}^{{x}} \:=\:\mathrm{8}\:{is}\:{symmetric} \\ $$$${about}\:{y}\:=\:{x} \\ $$

Commented by mr W last updated on 05/Aug/23

it′s obvious. when you replace x with  y and y with x, do you get the same?

$${it}'{s}\:{obvious}.\:{when}\:{you}\:{replace}\:{x}\:{with} \\ $$$${y}\:{and}\:{y}\:{with}\:{x},\:{do}\:{you}\:{get}\:{the}\:{same}? \\ $$

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