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Question Number 985 by tera last updated on 13/May/15
jika fungsi f(x)= px^2 −(p+1)x−6  mencapai nilai tertingi untuk   x=−1. maka nilai p=.....℧
$${jika}\:{fungsi}\:{f}\left({x}\right)=\:{px}^{\mathrm{2}} −\left({p}+\mathrm{1}\right){x}−\mathrm{6} \\ $$$${mencapai}\:{nilai}\:{tertingi}\:{untuk}\: \\ $$$${x}=−\mathrm{1}.\:{maka}\:{nilai}\:{p}=…..\mho \\ $$$$ \\ $$
Answered by prakash jain last updated on 13/May/15
f ′(x)=2px−(p+1), f ′′(x)=2p  f ′(−1)=0⇒−2p−p−1=0⇒p=−(1/3)  If p=−(1/3), f ′(−1)=0, f ′′(−1)=−(2/3)<0
$${f}\:'\left({x}\right)=\mathrm{2}{px}−\left({p}+\mathrm{1}\right),\:{f}\:''\left({x}\right)=\mathrm{2}{p} \\ $$$${f}\:'\left(−\mathrm{1}\right)=\mathrm{0}\Rightarrow−\mathrm{2}{p}−{p}−\mathrm{1}=\mathrm{0}\Rightarrow{p}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{If}\:{p}=−\frac{\mathrm{1}}{\mathrm{3}},\:{f}\:'\left(−\mathrm{1}\right)=\mathrm{0},\:{f}\:''\left(−\mathrm{1}\right)=−\frac{\mathrm{2}}{\mathrm{3}}<\mathrm{0} \\ $$

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