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anonymous

  • one year ago

Some help would be lovely !!:D Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x)=(x^2–16)/(x−2) , with x ≠ 2. A)On what intervals is the graph of g concave down? Justify your answer B)Write an equation for the tangent line to the graph of g at the point where x = 3.

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  1. anonymous
    • one year ago
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    @dan815

  2. jim_thompson5910
    • one year ago
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    are you able to find g '' (x) ?

  3. anonymous
    • one year ago
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    yes i got \[(x^2 -4x + 16)/((x-2)^2)\]

  4. triciaal
    • one year ago
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    |dw:1438285855095:dw|

  5. anonymous
    • one year ago
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    yep yep did that got plus or minus 4 for g'(x)

  6. jim_thompson5910
    • one year ago
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    If you complete the square for the numerator, you get this x^2 - 4x + 16 x^2 - 4x + 4 - 4 + 16 (x^2 - 4x + 4) + 12 (x-2)^2 + 12 So this means that x^2 - 4x + 16 is always positive. The same applies to (x-2)^2. Overall, g '' (x) is always positive

  7. jim_thompson5910
    • one year ago
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    since g '' (x) is never negative, it's impossible for g to be concave down

  8. anonymous
    • one year ago
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    okay so if g''(x) is positive then g(x) is concave up?

  9. jim_thompson5910
    • one year ago
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    yeah g is concave up over all of g's domain

  10. anonymous
    • one year ago
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    okay and as far as i can get on part b is getting the slope ... wich is -7

  11. anonymous
    • one year ago
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    so now i need a point on the tangent line right?

  12. jim_thompson5910
    • one year ago
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    g(3) = 4 which means (3,4) lies on the g(x) function use m = -7 x = 3 y = 4 with y = mx+b and solve for b

  13. anonymous
    • one year ago
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    4 = -7(3) + b 25 = b

  14. jim_thompson5910
    • one year ago
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    good

  15. anonymous
    • one year ago
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    so y = -7x + 25?

  16. jim_thompson5910
    • one year ago
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    yes

  17. anonymous
    • one year ago
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    okay great thank you very much !!!

  18. jim_thompson5910
    • one year ago
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    glad to be of help

  19. triciaal
    • one year ago
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    |dw:1438286497814:dw|

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