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Question Number 33036 by rahul 19 last updated on 09/Apr/18
![If range of f(x)= (x+1)(x+2)(x+3)(x+4)+5 x∈ [−6,6] is [a,b] ,a,b∈N, find a+b ?](Q33036.png)
$${If}\:{range}\:{of}\: \\ $$$${f}\left({x}\right)=\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)+\mathrm{5} \\ $$$${x}\in\:\left[−\mathrm{6},\mathrm{6}\right]\:{is}\:\left[{a},{b}\right]\:,{a},{b}\in{N},\:{find}\:{a}+{b}\:? \\ $$
Answered by MJS last updated on 09/Apr/18
![we need the minimum and maximum of f(x) in the given interval f(x)=x^4 +10x^3 +35x^2 +50x+29 f′(x)=4x^3 +30x^2 +70x+50= =2(2x+5)(x^2 +5x+5) f′(x)=0 x_1 =−(5/2)=−2.5 x_2 =−(5/2)−((√5)/2)≈−3.618 x_3 =−(5/2)+((√5)/2)≈−1.382 f′′(x)=12x^2 +60x+70 f′′(x_1 )<0 ⇒ x_1 is local maximum f′′(x_2 )>0 ⇒ x_2 is local minimum f′′(x_3 )>0 ⇒ x_3 is local minimum f(x_1 )=((89)/(16))=5.5625<f(−6)<f(6) f(x_2 )=f(x_3 )=4 max(f(x))=f(6)=5045; x∈[−6;6] min(f(x))=f(−(5/2)±((√5)/2))=4; x∈R 5045+4=5049](Q33057.png)
$$\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and} \\ $$$$\mathrm{maximum}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{interval} \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{35}{x}^{\mathrm{2}} +\mathrm{50}{x}+\mathrm{29} \\ $$$${f}'\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} +\mathrm{30}{x}^{\mathrm{2}} +\mathrm{70}{x}+\mathrm{50}= \\ $$$$=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{5}\right)\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right) \\ $$$${f}'\left({x}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{5}}{\mathrm{2}}=−\mathrm{2}.\mathrm{5} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{3}.\mathrm{618} \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{1}.\mathrm{382} \\ $$$${f}''\left({x}\right)=\mathrm{12}{x}^{\mathrm{2}} +\mathrm{60}{x}+\mathrm{70} \\ $$$${f}''\left({x}_{\mathrm{1}} \right)<\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{local}\:\mathrm{maximum} \\ $$$${f}''\left({x}_{\mathrm{2}} \right)>\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{local}\:\mathrm{minimum} \\ $$$${f}''\left({x}_{\mathrm{3}} \right)>\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{local}\:\mathrm{minimum} \\ $$$${f}\left({x}_{\mathrm{1}} \right)=\frac{\mathrm{89}}{\mathrm{16}}=\mathrm{5}.\mathrm{5625}<{f}\left(−\mathrm{6}\right)<{f}\left(\mathrm{6}\right) \\ $$$${f}\left({x}_{\mathrm{2}} \right)={f}\left({x}_{\mathrm{3}} \right)=\mathrm{4} \\ $$$$ \\ $$$${max}\left({f}\left({x}\right)\right)={f}\left(\mathrm{6}\right)=\mathrm{5045};\:{x}\in\left[−\mathrm{6};\mathrm{6}\right] \\ $$$${min}\left({f}\left({x}\right)\right)={f}\left(−\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\mathrm{4};\:{x}\in\mathbb{R} \\ $$$$\mathrm{5045}+\mathrm{4}=\mathrm{5049} \\ $$
Commented by rahul 19 last updated on 09/Apr/18
![i am not able to understand last 2 lines; what if 5.5625> f(6) ? and if it lies in between f(−6) , f(6)? similarly we have not put f(−6) as it does not give min. value ? why x∈[−6,6] andx∈ R othertime ?](Q33061.png)
$${i}\:{am}\:{not}\:{able}\:{to}\:{understand}\:{last}\:\mathrm{2}\:{lines}; \\ $$$${what}\:{if}\:\mathrm{5}.\mathrm{5625}>\:{f}\left(\mathrm{6}\right)\:?\:{and}\:{if}\:{it}\:{lies} \\ $$$${in}\:{between}\:{f}\left(−\mathrm{6}\right)\:,\:{f}\left(\mathrm{6}\right)? \\ $$$${similarly}\:{we}\:{have}\:{not}\:{put}\:{f}\left(−\mathrm{6}\right)\: \\ $$$${as}\:{it}\:{does}\:{not}\:{give}\:{min}.\:{value}\:? \\ $$$${why}\:{x}\in\left[−\mathrm{6},\mathrm{6}\right]\:{andx}\in\:{R}\:{othertime}\:? \\ $$
Commented by MJS last updated on 09/Apr/18
![in this case, f(6)=5045 is the maximum for x∈[−6;6] we found local maximum at x=−(5/2) which could be the maximum in [−6;6] but it′s not because f(−(5/2))=((89)/(16))<5045 so the maximum of f(x) im this interval is 5045 the minimum is 4, it′s even the absolute minimum of f(x) for x∈R ⇒ it′s also the minimum within [−6;6]](Q33063.png)
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case},\:{f}\left(\mathrm{6}\right)=\mathrm{5045}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{for}\:{x}\in\left[−\mathrm{6};\mathrm{6}\right] \\ $$$$\mathrm{we}\:\mathrm{found}\:\mathrm{local}\:\mathrm{maximum}\:\mathrm{at} \\ $$$${x}=−\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{which}\:{could}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{in}\:\left[−\mathrm{6};\mathrm{6}\right]\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not} \\ $$$$\mathrm{because}\:{f}\left(−\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{\mathrm{89}}{\mathrm{16}}<\mathrm{5045} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{im}\:\mathrm{this} \\ $$$$\mathrm{interval}\:\mathrm{is}\:\mathrm{5045} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{4},\:\mathrm{it}'\mathrm{s}\:\mathrm{even}\:\mathrm{the} \\ $$$$\mathrm{absolute}\:\mathrm{minimum}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{for} \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{within}\:\left[−\mathrm{6};\mathrm{6}\right] \\ $$
Commented by rahul 19 last updated on 09/Apr/18
$${thank}\:{u}\:{so}\:{much}\:{sir}\:! \\ $$
Commented by MJS last updated on 10/Apr/18
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$