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nickersia
Tangent at the point of inflection question.
Part d, help anyone? Thanks :)
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.
Everything is on the picture, I've done parts a,b and c, and I don't know how to do part d
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?
\[y-y _{1}=m(x-x _{1})\] \[y- (\frac{ 2 }{ 4 }-1+\ln \frac{ 2 }{ 4 }) = \frac{ x _{a} -2}{ x _{a} ^{2} }(x-4)\] where Xa is 4 because point B is (4,0.193)
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
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?
\[m = \frac{ x-2 }{ x ^{2} }\] where x is 4, so it gives me 0.125
Because \[\frac{ dy }{ dx } = m\]
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.
\[y-0.193 = (?)*(x - ?)\]
the second derivative must equal zero to be an inflection point
@nincompoop, are you saying that the inflection point is not (4,0.193)?
I didn't look at his solution, but that is what to bear in mind when looking for the inflection point.
@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.
\[y - (-\frac{ 1 }{2 }+\ln2) = \frac{ 1 }{ 8 }(x-4)\] . . . \[x-8y+8(\ln2-1)=0\] The problem was that I didn't changed the slope with 1/8 at the beginning so it got more and unnecessary complicated.
@nickersia: Congrats! Again, I admire the neatness and precision of your work.
@nincompoop yes, I used that in part b) Thank you @mathmale ! :)
Thanks for the medal and for becoming a fan of mine! Hope to work with you again soon.
You're welcome, me too! Good luck :)