SỬ DỤNG ĐẠO HÀM ĐỂ TÌM GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ ITÓM TẮT LÍ THUYẾT ≤ ∀ ∈ ∃ ∈ = ∈ = ≥ ∀ ∈ ∃ ∈ = ∈ = IIPHƯƠNG PHÁP GIẢI TOÁN !""#$%&"' ()*+,-."/'01+23'01+ 345( / 6 / 7"8& 9:$'."';-."/'8%<60- +:-."/'0=<%(>+<&#<1" <&0+"+?>"<10+"@>;5 -."/' ()*+,&A/'B6C0=D& 1?&&045 A / B 6 A / B E1".'(01$%,"".;'F ++:$'."';&A/'B8%<60-+: ! "#$%& ' '(' )*"+,&-./ "0*"'* "'* "'('* "'*" 1" ( ) ( ) ( ) ( ) { } G H A / B = . ( ) ( ) ( ) ( ) { } G H A / B = IIIBÀI T)P ÁP DỤNG B*i 1 I9I;J:$K"LLM"- K H K N KO = − − + &A2P/PB 2 0 H K L N = − − H G Q K L N Q K = − = ⇔ − − = ⇔ = ."(R=&A2P/PB *2P)2PG*2G)PQ*K)S*P)GO T:( A P/PB PQ − = 6 A P/PB PG − = − B*i 2 I9I;J:$K%"LLM"- H = − & / H H π π − 2 0 HH G = − H H G L Q HH G Q H H H H L 3 2 π π π π = + = ⇔ − = ⇔ = ⇔ = − + & / H H π π − $ U"0"V+H)W K π L π ⇔ = ± K K H H H H L H L L H L π π π π π π π π − = = − − = − + = − ÷ ÷ ÷ ÷ T:(() H π () H π B*i 3"+?>"@>;J:$K'"LLM"- H K H = − + &A2GQ/GQB J+ 0 [ ] [ ] [ ] H H K H GQ/G H/GQ K H G/H − + ∈ − ∪ = − + − ∈ [ ] [ ] [ ] H K GQ/G H/GQ H K G/H − ∈ − ∪ = − + ∈ J."' 2GQG K H HGQ 2Q 2 ( GKH G P XH QQ 6'."'(R";(+AQ/GKHB A / B GKH = A / B Q = B*i 4 "+?>"@>; ( ) G H G H = + + + 2 0 ( ) H L P H G H H H = + + + + + + E#) H ≤ H G H ⇒ = + Y0 ( ) H H H H H G H K = = + − + + 6? H ≤ 0 H HH P H + + 6? G K G K H/ / H H H − − − + ∈ − ∪ ") H P S− + 6? G K G K / H H − − − + ∈ G K G K S G H/ / H H H G K G K S / H H − − − + + ∈ − ∪ = − − − + − ∈ J."' H− G K H − − Q G K H − + H ( ) 2Q2 " P H K− 6'."'(R";"+ H P H K/P H G − + *ZQ ( ) * ) P H K ) K G− − / ( ) H * )P H G+ B*i 5 I[ G H +"V;$ U" H H H GH GH L P Q − + − + = K K G H + E&I 'E&I9 2! E1$ U"0"V H Q P GH H K/ H H/H K ⇔ ∆ ≥ ⇔ ≤ ≤ ⇔ ∈ − − ∪ \+]T2\ G H H + = H G H H G GH P GH = − + ÷ 0 ( ) K K K G H G H G H G H K K H H 4 = + = + − + = − H P H G + P H K+ S H P H G− H G K Q Q H H 4 = + > ∀ ≠ J."' 23 H K− 2HQH H K 3 4 ^ G P 0^+?>) K K P ^@>) K K P − B*i 6 I96I; L L P P G G + + = + + 2 0 H H K H H P G H H H − = − E# [ ] H H Q/G5 5= ∈ 0 ( ) K H S K K S P G S H H S H H 5 5 5 4 5 5 5 5 − − − = = = = − − − M( ( ) ( ) [ ] H S Q Q/G H S 4 5 5 5 − = < ∀ ∈ − ^_"'&AQ/GB0 I9;^_)^Q)GI9;^_)^G) O L B*i 7 $ U" H H H H G P K Q + + + + + = I[ G H +"VI9; G H G H H 6 = − + 2! E1$ U"0"V H Q L O Q O G ⇔ ∆ ≥ ⇔ + + ≤ ⇔ − ≤ ≤ − \+]T`\0 ( ) H G H G H P K G H + + + = − + = K K P − G P K K P ( ) ( ) H H H P K G G H G S X S X H H H 6 + + = + + = + + = − + + &A2O/2GB 0 P6 = − − &A2O/2GBJ."' 2O2P2G 6 Q2 a N H PQ T:(I9;a'b" N H 789#:;<':<<)=> B*i 8 H H G + = ; H H H H H G ? + = + + 2 H H H H H H H H H H H H H H G H H K H ? + + + = = = + + + + + + + ()QM)Q (cQ.d6e;M H 0 ( ) H H H G H G K H G K H G ? + ÷ + = = + + + + ÷ 6? = 0 ( ) H H H H K L G K H G ? − + + = + + G H K L K L Q K K ? − − − + = ⇔ = = f#- ( ) H H G + Q K H G →∞ + = + + J."' 23 G H 3 ? 2QQ2 M Q H L H + H L H − Q 6'."'( H L H ? − = H L H ? + = B*i 9 "@>; P P H H P P H H 4 = + − + + + ÷ T?'cQ 2 E# 5 = + H Q5 ≥ ∀ ≠ Y0 ( ) ( ) H H H P H H H H O P4 5 5 5 5 5 5= − − − − + = − + + K P GQ G4 5 = − + H GH GQ4 5= − M( _ZH ( ) ( ) Q H GK Q4 4 5 4⇒ > ⇒ ≥ = > _g2H ( ) ( ) Q H GG4 4 5 4⇒ > ⇒ ≥ − = − J."' _ 232HH3 4 2 ^ 33 2HH I@>;^)2H B*i 10 (ZQ6()GI9I; G G ? = + + + 2 h".( G Q P ≤ ≤ 0 H H G ? + + + = + + + ( ) H H G G + − + = + + + H H H − = + E#() G Q P ≤ ≤ ( H H H ? − = + ( ) H L Q H H ? − = ≤ ∀ ≠ − + J."' 6'."'(I9;M'b"G/I;M'b" H K B*i 11 ZQ <(55(ZG ()K "@>+?>;'1Fi) K H( H K H P(2O 2 ZQ ()K)j()K20)jQggH (ZG i) K H `OH( H H( H ) K H `OH( H ) K H `OGS _k*) K H `OGS/ AQ/HB∀ ∈ *l)K H H`O/*l)Qm)j G O K = = − / O K = − +& 0*Q)GS/*G)GO/*H)HQ T:(I9;i'b"HQ/I;i'b"GO J:$<". JGT?"; H O P = − + + 0I+?UGn JH$o1"+?>; H 9 = + + &A2G/GB+'k> JK(jQ6()G I; G G ? = + − − JPI.d H H G Q 9 9 + + = 0"V G H $cQM) G P H P @> . SỬ DỤNG ĐẠO HÀM ĐỂ TÌM GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ ITÓM TẮT LÍ THUYẾT . = . ( ) ( ) ( ) ( ) { } G H A / B = IIIBÀI T)P ÁP DỤNG B*i 1 I9I;J:$K"LLM"- K H K