anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
are you able to find g '' (x) ?
anonymous
  • anonymous
yes i got \[(x^2 -4x + 16)/((x-2)^2)\]

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triciaal
  • triciaal
|dw:1438285855095:dw|
anonymous
  • anonymous
yep yep did that got plus or minus 4 for g'(x)
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
since g '' (x) is never negative, it's impossible for g to be concave down
anonymous
  • anonymous
okay so if g''(x) is positive then g(x) is concave up?
jim_thompson5910
  • jim_thompson5910
yeah g is concave up over all of g's domain
anonymous
  • anonymous
okay and as far as i can get on part b is getting the slope ... wich is -7
anonymous
  • anonymous
so now i need a point on the tangent line right?
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
4 = -7(3) + b 25 = b
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
so y = -7x + 25?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
okay great thank you very much !!!
jim_thompson5910
  • jim_thompson5910
glad to be of help
triciaal
  • triciaal
|dw:1438286497814:dw|

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