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anonymous

  • one year ago

Consider the function f(x) = -5 x - 2 and find the following: a) The average rate of change between the points (-1, f(-1) ) and (1, f( 1 ) ) . b) The average rate of change between the points (a, f(a) ) and (b, f(b) ) . c) The average rate of change between the points (x, f(x) ) and (x+h, f(x+h) ) .

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  1. Jhannybean
    • one year ago
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    Average rate of change : \(A(x) = \dfrac{f(x)-f(a)}{x-a}\)

  2. anonymous
    • one year ago
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    where do i find a if like in a) i plug in -1 to the f(x) which is -5(-1)-2

  3. anonymous
    • one year ago
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    ohh

  4. Jhannybean
    • one year ago
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    \[A(x) = \frac{f(-1) - f(1)}{-1 -(1)}\]

  5. Jhannybean
    • one year ago
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    Think you can handle the rest?

  6. anonymous
    • one year ago
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    i got, thank you!

  7. Jhannybean
    • one year ago
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    sweet.

  8. Jhannybean
    • one year ago
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    Just remember that the 2 points represent 2 coordinate points, and you're trying to find the slope \[A(x) = \frac{f(x)-f(a)}{x-a} \iff m = \frac{y_2 -y_1}{x_2-x_1}\]

  9. Jhannybean
    • one year ago
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    When you think about it this way, it's easier to identify which points subtract what, I guess

  10. anonymous
    • one year ago
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    ohhh thats very helpful, thanks

  11. Jhannybean
    • one year ago
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    mmhmm, no problem

  12. anonymous
    • one year ago
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    wait can you check my answers cause for a, i got -1 and for b i got a^2-b^2-5. I think i plugged in wrong

  13. Jhannybean
    • one year ago
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    Think I might have stated the wrong values of x, but this is how i solved it. \[f(a)=f(-1) = 3~,~f(x)= f(1) = -7\]\[m = \frac{f(x)-f(a)}{x-a} = \frac{-7-(3)}{1-(-1)} =-\frac{10}{2}=-5 \]

  14. anonymous
    • one year ago
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    oh wow okay i was looking at a similar but different problem...I think i can do it now, thanks!

  15. Astrophysics
    • one year ago
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    Yeah it's basically \[\frac{ y_2-y_1 }{ x_2-x_1 }\] notice it's the slope formula @Jhannybean it's better to write the equation as \[m = \frac{ f(b)-f(a) }{ b-a }\] to avoid confusion

  16. Jhannybean
    • one year ago
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    And i figured I switched it because when you solve the last one with \((x~,~ f(x))\) and \((x+h,~f(x+h))\) you will get the definition of a derivative once you take the limit of it, but that's for later i guess. Therefore I knew i made a mistake.

  17. anonymous
    • one year ago
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    how would i do c?

  18. Jhannybean
    • one year ago
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    so you have (x, f(x) ) and (x+h, f(x+h) ) Which translates into \[m= \frac{f(x+h)-f(x)}{(x+h)-x}\]

  19. anonymous
    • one year ago
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    oh i just leave the f(x)

  20. Jhannybean
    • one year ago
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    Hm? Nothing changes in the numerator, it's only the denominator that you're concerned wtih.

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