anonymous
  • anonymous
Values of x and g(x) are given in the table below. For what value of x in the table is g'(x) closest to 3?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
I have attempted to compare each point separately as well as use the difference quotient but I can't seem to find the right answer.
eSpeX
  • eSpeX
@ganeshie8 might be a good one to ask here. I was trying a similar method as you and don't like where it was going.

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anonymous
  • anonymous
Okay, thank you! I've been stuck on this one for a few days on my online homework.
eSpeX
  • eSpeX
Okay, how about this. Think about "slope" (your derivative) is equal to your rise (difference in y) over run (difference in x).
anonymous
  • anonymous
Okay so 3 = (y1-y2)/(x1-x2)
eSpeX
  • eSpeX
Yes, or approximately 3, since they do tell you that it is close to 3
anonymous
  • anonymous
I think I am seeing 6.4 as the answer if I use points on either side of it.
eSpeX
  • eSpeX
That was not what I got, are you doing \(g'(x)=\frac{y_2-y_1}{x_2-x_1}\)?
eSpeX
  • eSpeX
Technically I guess it is \(g'(x)=\frac{f(x)_2-f(x)_1}{x_2-x_1}\)
anonymous
  • anonymous
Yes, I used the points (6.9, 11.7) and (5.9, 8.1).
eSpeX
  • eSpeX
From 6.1 to 6.7 I am getting 1.2, 8.1 to 6.7 I am getting 2.8
anonymous
  • anonymous
I got that for those too. So would I say that the x-value is 5.9? As using 6.4 and 5.9 gives 3.2 as well?
eSpeX
  • eSpeX
I suppose it may be a difference of how you evaluate the instructions. I would say that "closest to 3" is defined as not going over and so I would have put the value to 5.9 for 'x' over 6.4.
anonymous
  • anonymous
Okay. Thank you for your help. I think I will try to ask my instructor for what she means in the wording of the question but I think you are right about the approach of not going over 3.
eSpeX
  • eSpeX
You're welcome, good luck.

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