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Pertanyaan

J i k a space k u r v a space space y equals open parentheses x squared minus a close parentheses open parentheses 2 x plus b close parentheses cubed space space t u r u n space p a d a space i n t e r v a l space space minus 1 less than x less than 2 over 5 space space m a k a space n i l a i space a b space equals space horizontal ellipsis.

  1. negative 3

  2. negative 2

  3. 1

  4. 2

  5. 3

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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Pembahasan

Dari persamaan (i) dan (ii) sama, sehingga diperoleh :

y equals open parentheses x squared minus a close parentheses open parentheses 2 x plus b close parentheses cubed space space b e n t u k n y a space space y equals u. v space rightwards double arrow y to the power of apostrophe equals u to the power of apostrophe. v plus u. v apostrophe y equals open parentheses 2 x close parentheses open parentheses 2 x plus b close parentheses cubed plus open parentheses x squared minus a close parentheses open parentheses 3 close parentheses open parentheses 2 x plus b close parentheses squared open parentheses 2 close parentheses equals open parentheses 2 x close parentheses open parentheses 2 x plus b close parentheses open parentheses 2 x plus b close parentheses squared plus open parentheses x squared minus a close parentheses open parentheses 2 x plus b close parentheses squared open parentheses 6 close parentheses equals open parentheses 2 x plus b close parentheses squared open curly brackets open parentheses 2 x close parentheses open parentheses 2 x plus b close parentheses plus open parentheses x squared minus a close parentheses open parentheses 6 close parentheses close curly brackets

equals open parentheses 2 x plus b close parentheses left parenthesis 2 x plus b right parenthesis to the power of space open curly brackets 4 x squared plus 2 b x plus 6 x squared minus 6 a close curly brackets equals left parenthesis 2 right parenthesis open parentheses 4 x squared plus 4 b x plus b squared close parentheses to the power of space left curly bracket 5 x squared plus b x minus 3 a right curly bracket equals 2 open parentheses 2 x plus b close parentheses squared open parentheses 5 x squared plus b x minus 3 a close parentheses

 

K u r v a space y space m e r u p a k a n space f u n g s i space t u r u n space j i k a space space y to the power of apostrophe less than 0 space space s e h i n g g a space space left parenthesis 2 right parenthesis open parentheses 4 x squared plus 4 b x plus b squared close parentheses to the power of space open curly brackets 5 x squared plus b x minus 3 a close curly brackets less than 0

K a r e n a space n i l a i space space 2 open parentheses 2 x plus b close parentheses squared space space s e l a l u space p o s i t i f comma space m a k a space space open curly brackets 5 x squared plus b x minus 3 a close curly brackets less than 0 space space horizontal ellipsis space left parenthesis i right parenthesis

 

I n t e r v a l space space minus 1 less than x less than 2 over 5 space comma space m a k a space colon open parentheses x plus 1 close parentheses open parentheses 5 x minus 2 close parentheses less than 0 5 x squared plus 3 x minus 2 less than 0 space space horizontal ellipsis space left parenthesis i i right parenthesis

Dari persamaan (i) dan (ii) sama, sehingga diperoleh :

b space equals space 3 space space d a n space minus 3 a equals negative 2 left right double arrow a equals 2 over 3 J a d i comma space n i l a i space space a b equals open parentheses 2 over 3 close parentheses 3 equals 2

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Nilai maksmimum dan minimum untuk fungsi f x =2 x 2 - 8 x pada interval − 1 ≤ x ≤ 4 - 1≤ x ≤4 adalah ...

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