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anonymous

  • 5 years ago

Let f(x)={4x−1, if x≤3...... .............−5x+b, if x>3 If f(x) is a function which is continuous everywhere, then we must have b=

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  1. myininaya
    • 5 years ago
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    do the left limit and right limit for x=3 and set them equal and solve for b

  2. anonymous
    • 5 years ago
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    ohh i see

  3. anonymous
    • 5 years ago
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    how would i go about solving this one...same way? For what value of the constant c is the function f continuous on (−∞,∞) where f(x)=...... cx+8 if x∈(−∞,2] ....... cx^2−8 if x∈(2,∞)......

  4. myininaya
    • 5 years ago
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    yes you look at the left limit and right limit for x=2

  5. myininaya
    • 5 years ago
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    and set them equal

  6. anonymous
    • 5 years ago
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    hey so for the first problem...i used 4(3)-1 = 11 and -5(3.1)+b = -15.5 + b. So I set -15.5 + b = 11 and I did the algebra and ended up with -4.5. what did i do wrong?

  7. myininaya
    • 5 years ago
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    \[4(3)-1=-5(3)+b\]

  8. anonymous
    • 5 years ago
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    yeah shouldnt u get a -4.5 after you do the algebra

  9. myininaya
    • 5 years ago
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    i don't know i can perform the operations if you want me too

  10. myininaya
    • 5 years ago
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    4(3)=12 12-1=11 -5(3)=-15 so we have 11=-15+b By adding 15 on both sides we get 26=b

  11. anonymous
    • 5 years ago
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    oh snap i did it backward haha thanks

  12. anonymous
    • 5 years ago
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    can i use 1.5 and 3 for the second question

  13. myininaya
    • 5 years ago
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    i don't understand

  14. anonymous
    • 5 years ago
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    how would i go about solving this one...same way? For what value of the constant c is the function f continuous on (−∞,∞) where f(x)=...... cx+8 if x∈(−∞,2] ....... cx^2−8 if x∈(2,∞)...... once again i have to approach from both sides...if im approaching from left can I use x=1.5 and right x=3?

  15. myininaya
    • 5 years ago
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    what no just before we plugged in 3 for both sides here we do 2

  16. myininaya
    • 5 years ago
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    \[c(2)+8=c(2)^2-8\]

  17. myininaya
    • 5 years ago
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    now solve for c

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