# Math

posted by .

Find the coordinates of the turning point and determine wether it is minimum or maximum.
y=xlnx-2x

The answer in the book says the co-ordinates are (e,-e), the closest I have come is (1/lnx,-1/lnx) which works if 1/lnx=e, but I don't think that it does. This question is in relation to derivatives.

## Similar Questions

1. ### math

how would you solve for y in this problem: ln(y-1)-ln2 = x +lnx ln(y-1)-ln2 = x +lnx Solve for y... ln[(y-1)/2]]=x + lnx take the antilog of each side (y-1)/2= e^(x+lnx) solve for y.
2. ### calc

integral of 1 to e^4 dx/x(1+lnx) 1+lnx = a => dx/x = da dx/ x(1+lnx) = da/a int of da/a = In(a) =>int of dx/ x(1+lnx) = In(1+In(x)) + C from 1 to e^4 for int of dx/ x(1+lnx) = In(1+In(1)) - In(1+In(e^4)) In(1+In(1)) - In(1+In(e^4)) …
3. ### L'Hopital's rule

Find lim x->1+ of [(1/(x-1))-(1/lnx)]. Here is my work... =(lnx-(x-1)) / ((x-1)(lnx)) =(lnx-1) / (lnx+ (x+1)/x) This becomes(1/x) / ((1/x)+(1/x^2)) which becomes 1/ (1/x^2) This equals 1/2. I understand the answer has to be -1/2, …
4. ### College Calc

Problem: y=sqrtx ^1/x. Find DY/DX. ln(f(x))=lnx^1/2x or lnx to the 1/2x 1/2x lnx product rule (1/4x^2)(lnx)+(1/x)(1/2x) is the product rule correct?
5. ### Calculus 1

Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36) So far this is what I've gotten: y=(x^6/6)(lnx)-(x^6/36) y=(1/6)x^6(lnx)-(1/36)x^6 y'=(1/6)x^5(1/x)+lnx(x^5)-(1/6)x^5 What do I do now?
6. ### calculus

sorry to ask a second question so soon, but i'm just not getting this one. if f(x)= 3x lnx, then f'(x)=?
7. ### Calculus

Find the derivatives, dy/dx for the following functions. a. x^2y^4 = lnx + 3y b. y = ln(2x^2 + 1) c. y = ln(ãx) d. y = lnx/x e. y = x^2 lnx f. y ß log_2⁡(3x)
8. ### Calculus

what is the limit as h approaches 0 of ((x+h)^pi-x^pi)/h?
9. ### calculus

Find the minimum distance between the curves y=e^x and y=lnx. Hint: Use the fact that e^x and lnx are inverse relationships. I have no idea where to start. Thanks!
10. ### Calculus

Find the minimum distance between the curves y=e^x and y=lnx. Hint: Use the fact that e^x and lnx are inverse relationships. I have no idea where to start. Thanks! calculus - Reiny, Monday, January 18, 2016 at 9:02pm to get you going: …

More Similar Questions