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Thread: Find a for f(x+6)=2x+a where f(a)=20

  1. #1
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    Find a for f(x+6)=2x+a where f(a)=20

    I haven't a clue about how to do this. Its my last question and I've been looking at it for almost a half hour now with no result. Any help appreciated...thanks!
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    Quote Originally Posted by GeeDee View Post
    I haven't a clue about how to do this. Its my last question and I've been looking at it for almost a half hour now with no result. Any help appreciated...thanks!
    I think it's

    f(x+6)=2x+a

    f(x+6)=2(x+6)+20

    0=2x+12+20

    x=-16

    (-16+6)=2(-16)+a

    a=22

    I just taking a guess here
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  3. #3
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    Thanks....I just actually tried doing that before I read your post. But apparently the answer in the book tells me a=8/3 ?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by GeeDee View Post
    I haven't a clue about how to do this. Its my last question and I've been looking at it for almost a half hour now with no result. Any help appreciated...thanks!
    If your book told you the answer is a = 8/3, then the book is wrong.

    You have
    $\displaystyle f(x + 6) = 2x + a$
    and you want
    $\displaystyle f(a) = 20$.

    Well, if $\displaystyle f(x + 6) = 2x + a$ then

    Let $\displaystyle y = x + 6 \implies x = y - 6$ so
    $\displaystyle f(y) = 2(y - 6) + a$

    $\displaystyle f(y) = 2y - 12 + a$

    We want $\displaystyle f(a) = 20$, so
    $\displaystyle f(a) = 2a - 12 + a = 20$

    $\displaystyle 3a - 12 = 20$

    $\displaystyle 3a = 32$

    $\displaystyle a = \frac{32}{3}$

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    If your book told you the answer is a = 8/3, then the book is wrong.

    You have
    $\displaystyle f(x + 6) = 2x + a$
    and you want
    $\displaystyle f(a) = 20$.

    Well, if $\displaystyle f(x + 6) = 2x + a$ then

    Let $\displaystyle y = x + 6 \implies x = y - 6$ so
    $\displaystyle f(y) = 2(y - 6) + a$

    $\displaystyle f(y) = 2y - 12 + a$

    We want $\displaystyle f(a) = 20$, so
    $\displaystyle f(a) = 2a - 12 + a = 20$

    $\displaystyle 3a - 12 = 20$

    $\displaystyle 3a = 32$

    $\displaystyle a = \frac{32}{3}$

    -Dan
    i thought as much. i saw no need to introduce a new variable though.

    $\displaystyle f(x + 6) = 2x + a = 2x + 12 - 12 + a = 2(x + 6) - 12 + a$

    and then, continue as you did
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    i thought as much. i saw no need to introduce a new variable though.

    $\displaystyle f(x + 6) = 2x + a = 2x + 12 - 12 + a = 2(x + 6) - 12 + a$

    and then, continue as you did
    I was trying not to confuse the original poster with the "x + 6 - 6" thing. Maybe good, maybe not.

    -Dan
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    I was trying not to confuse the original poster with the "x + 6 - 6" thing. Maybe good, maybe not.

    -Dan
    yeah, you're right. that maneuver does confuse some people
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