anonymous
  • anonymous
Find the equation of tangent to the curve y = ln x at the point where x = a. Hence, find the equation of tangent to the curve which passes through the origin. Given that line y = mx intersects the curve the curve y = ln x at two distinct pts, write down an inequality for m.
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
what do we get for the derivative?
amistre64
  • amistre64
the derivative defines the slope at any given value of x; therfore the tangent line at any given points is defined as: y = f'(a) (x-a) + f(a)
amistre64
  • amistre64
then your next task seems to define the range of value for m such that lnx = mx has only 2 solutions

Looking for something else?

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

More answers

anonymous
  • anonymous
Yes and then???
amistre64
  • amistre64
and then you come up with the results ...
anonymous
  • anonymous
I still dont get the first part.
amistre64
  • amistre64
then ask questions about it; what dont you understand?
anonymous
  • anonymous
I differentiate ln x i get 1/x
amistre64
  • amistre64
good, and 1/x at any given x=a is just 1/a
anonymous
  • anonymous
Correct
amistre64
  • amistre64
y = 1/a (x-a) + ln(a) is the slope of the tangent line
amistre64
  • amistre64
lol, equation of the tangent line
anonymous
  • anonymous
*x-a + ln a * i see
amistre64
  • amistre64
when does this equation have an x and y intercept at the origin?
anonymous
  • anonymous
Lol i only know that method
amistre64
  • amistre64
in other words, for what value of a is (0,0) a good solution to it?
amistre64
  • amistre64
y = 1/a (x-a) + ln(a) 0 = 1/a (0-a) + ln(a) 0 = -1 + ln(a) 1 = ln(a) , when a=e
amistre64
  • amistre64
|dw:1377881456934:dw|
amistre64
  • amistre64
m=1/e ... just cant typ today
amistre64
  • amistre64
as long as m is between 0 and 1/e, the line from the origin should hit it in exactly 2 points
anonymous
  • anonymous
Can u give me a link to learn this concept?
amistre64
  • amistre64
sorry, my brain is all the resources I know of at the moment.

Looking for something else?

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