Where on the curve y = (4+x^2)^-1 does the tangent line have the greatest slope?
Have you considered the 1st Derivative to create a function describing the slope at each point?? Have you considered the 2nd derivative to find the extrema of this new function?
Yes. I took each derivative and got the first to be -2x(4+x^2)^-2 and the second to be (6x^2 - 8) / (4+x^2)^3
I thought I was to make the second derivative equal to zero, find the points, determine which is a max, and plug it into the derivative to determine the y value for the determined x value.
Okay, then why not just answer the question?
I did, but apparently I keep getting it wrong.
Why do you think your answer is incorrect?
The online program says its wrong
We use WileyPLUS
Okay. Your derivatives look good. Where else could it go wrong?
So I get that x = \[\sqrt{4/3}\], yeah?
That's good, as long as you get both + and -. Which is the greater slope?
Blah, it's the negative root. >_< I probably kept forgetting the sign
Very good. Now, we have \(x = -\dfrac{2}{3}\sqrt{3}\). How do we find y?
Why is x = to -2/3 sqrt3?
Standard simplification. \(x = -\dfrac{2}{\sqrt{3}} = -\sqrt{\dfrac{4}{3}}\) if you like.
Fair enough, I am quite terrible at roots
Gotta get up to speed on that!
Yeah, that and a lot of things. >_< Alright, y is..
Wait, so was the -2/3 a typo or am I missing another part?
The moment of truth. In which or our three functions will you substitute our x-value to determine the proper y-value?
First derivative, I want to find out when it is greatest
There where we wandered off! The question is "where on the curve is the greatest slope", not "what is the greatest slope". Please substitute into the original function to obtain the answer to the problem statement.
sigh. This chapter...
Hang in there. You are doing well. Just pay a little better attention.
3/16?
That's it. See, good work!
Thanks a lot for the help, its definitely much appreciated!
No worries. We're here to help. It's a great pleasure to see when it does help.
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