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Question Number 170566 by cortano1 last updated on 27/May/22

Find min value f(x)= (x+4)(x+5)(x+6)(x+7)

$$\:\:\mathrm{Find}\:\mathrm{min}\:\mathrm{value}\: \\ $$$$\:\:{f}\left({x}\right)=\:\left({x}+\mathrm{4}\right)\left({x}+\mathrm{5}\right)\left({x}+\mathrm{6}\right)\left({x}+\mathrm{7}\right) \\ $$

Answered by som(math1967) last updated on 27/May/22

(x+4)(x+7)(x+5)(x+6) (x^2 +11x+28)(x^2 +11x+30) (a+28)(a+30) [let a=x^2 +11x] a^2 +58a+840 (a+29)^2 −29^2 +840 (a+29)^2 −1 (a+29)^2 ≥0 (a+29)^2 −1≥−1 ∴(x+4)(x+5)(x+6)(x+7)≥−1 ∴ min value of f(x) is −1

$$\left({x}+\mathrm{4}\right)\left({x}+\mathrm{7}\right)\left({x}+\mathrm{5}\right)\left({x}+\mathrm{6}\right) \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{28}\right)\left({x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{30}\right) \\ $$$$\left({a}+\mathrm{28}\right)\left({a}+\mathrm{30}\right)\:\:\:\:\left[{let}\:{a}={x}^{\mathrm{2}} +\mathrm{11}{x}\right] \\ $$$${a}^{\mathrm{2}} +\mathrm{58}{a}+\mathrm{840} \\ $$$$\left({a}+\mathrm{29}\right)^{\mathrm{2}} −\mathrm{29}^{\mathrm{2}} +\mathrm{840} \\ $$$$\left({a}+\mathrm{29}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\left({a}+\mathrm{29}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({a}+\mathrm{29}\right)^{\mathrm{2}} −\mathrm{1}\geqslant−\mathrm{1} \\ $$$$\therefore\left({x}+\mathrm{4}\right)\left({x}+\mathrm{5}\right)\left({x}+\mathrm{6}\right)\left({x}+\mathrm{7}\right)\geqslant−\mathrm{1} \\ $$$$\therefore\:{min}\:{value}\:{of}\:\:{f}\left({x}\right)\:{is}\:−\mathrm{1} \\ $$

Answered by ajfour last updated on 27/May/22

let x+((11)/2)=t f(t)=(t−(3/2))(t−(1/2))(t+(1/2))(t+(3/2)) =(t^2 −(9/4))(t^2 −(1/4)) let t^2 −(5/4)=z f(z)=z^2 −1 f(z)∣_(min) =−1 when t=x+((11)/2)=±((√5)/2) x=((−11±(√5))/2)

$${let}\:\:\:{x}+\frac{\mathrm{11}}{\mathrm{2}}={t} \\ $$$${f}\left({t}\right)=\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\left({t}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${let}\:\:{t}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}={z} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} −\mathrm{1} \\ $$$${f}\left({z}\right)\mid_{{min}} =−\mathrm{1} \\ $$$${when}\:\:\:{t}={x}+\frac{\mathrm{11}}{\mathrm{2}}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\:{x}=\frac{−\mathrm{11}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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