anonymous
  • anonymous
Tangent at the point of inflection question.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
http://prntscr.com/356je6
anonymous
  • anonymous
Part d, help anyone? Thanks :)
mathmale
  • mathmale
Would you please start the necessary work, and share with us what you have done. also, please explain what it is that you need to know. Let's build upon what you do already know.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Everything is on the picture, I've done parts a,b and c, and I don't know how to do part d
mathmale
  • mathmale
Your work is very neat. I apologize for not having scrolled down to see what you'd already done. Let me look at Part D. What do you think you need to know to answer Part D correctly?
anonymous
  • anonymous
\[y-y _{1}=m(x-x _{1})\] \[y- (\frac{ 2 }{ 4 }-1+\ln \frac{ 2 }{ 4 }) = \frac{ x _{a} -2}{ x _{a} ^{2} }(x-4)\] where Xa is 4 because point B is (4,0.193)
anonymous
  • anonymous
y1 (this thing in the bracket) is equal to 0.193 which I got as a coordinate for B, and the slope of the tangent should be equal to derivative of y at point B
mathmale
  • mathmale
I'm going to assume that your point of inflection, (4,0.193), is correct. then \[x _{0}=4,y _{0}=0.193\] and you need only substitute these into the point-slope formula for the equation of a straight line. What is your assumed value for the slope, m?
anonymous
  • anonymous
\[m = \frac{ x-2 }{ x ^{2} }\] where x is 4, so it gives me 0.125
anonymous
  • anonymous
Because \[\frac{ dy }{ dx } = m\]
mathmale
  • mathmale
Using your formula for the slope, and calculating the slope that way, I get the same result calculating the slope differently. I, too, get m=0.125, which is the same as 1/8. I'd suggest you substitute these numerical results into the point slope form. Point of tangency is (4,0.193), as before, and slope is either 1/8 or 0.125.
mathmale
  • mathmale
\[y-0.193 = (?)*(x - ?)\]
nincompoop
  • nincompoop
the second derivative must equal zero to be an inflection point
mathmale
  • mathmale
@nincompoop, are you saying that the inflection point is not (4,0.193)?
nincompoop
  • nincompoop
I didn't look at his solution, but that is what to bear in mind when looking for the inflection point.
nincompoop
  • nincompoop
http://www.math.sc.edu/~diestelr/4.2Notes.pdf
anonymous
  • anonymous
I got it!
mathmale
  • mathmale
@nincompoop: If you haven't looked at Nick's solution, then commenting as you have is irrelevant and distracting. If you had found a mistake, then I would have wanted to hear from you.
anonymous
  • anonymous
\[y - (-\frac{ 1 }{2 }+\ln2) = \frac{ 1 }{ 8 }(x-4)\] . . . \[x-8y+8(\ln2-1)=0\] The problem was that I didn't changed the slope with 1/8 at the beginning so it got more and unnecessary complicated.
mathmale
  • mathmale
@nickersia: Congrats! Again, I admire the neatness and precision of your work.
anonymous
  • anonymous
@nincompoop yes, I used that in part b) Thank you @mathmale ! :)
mathmale
  • mathmale
Thanks for the medal and for becoming a fan of mine! Hope to work with you again soon.
anonymous
  • anonymous
You're welcome, me too! Good luck :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.