Finding all tangent lines through a point

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SUMMARY

The discussion focuses on finding all tangent lines of the function f(x) = x + 3/x that intersect the y-axis at 4. The point of tangency is defined as (a, a + 3/a), where the slope of the tangent line is calculated using the derivative f'(x) = 1 - 3/(x^2). The solution reveals that a = 3/2 is one point of tangency, but participants note that there should be a second point for x < 0, which is confirmed to not exist based on the graph of f(x).

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with the function f(x) = x + 3/x
  • Knowledge of tangent lines and their properties
  • Graphing skills to analyze function behavior
NEXT STEPS
  • Study the implications of the derivative f'(x) = 1 - 3/(x^2) in detail
  • Explore the graphical behavior of rational functions like f(x) = x + 3/x
  • Learn how to determine points of tangency for various functions
  • Investigate the conditions under which tangent lines can intersect a given point
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators looking for examples of rational function behavior.

hlin818
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Homework Statement



Find all tangent lines of the graph f(x)=x+3/x that have a y intercept of 4.

Homework Equations





The Attempt at a Solution



Assume a is the x coordinate of a point of tangency. Thus the point of tangency is (a, a+3/a). We know the tangent line must pass through (0,4) so the slope of the line must be (a+3/a-4)/(a-0).

f'(x)=1-3/(x^2)

Derivative of f at point a must equal the slope of the tangent line, i.e. we must have
f'(a)=1-3/(a^2)=(a+3/a-4)/(a-0)=m

Solving I get a=3/2. However, looking at the graph of f(x), it seems there should be two points where the tangent line passes through 4, the other one being on the part of the f(x) where x<0. Where did I go wrong?
 
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hlin818 said:

Homework Statement



Find all tangent lines of the graph f(x)=x+3/x that have a y intercept of 4.

Homework Equations





The Attempt at a Solution



Assume a is the x coordinate of a point of tangency. Thus the point of tangency is (a, a+3/a). We know the tangent line must pass through (0,4) so the slope of the line must be (a+3/a-4)/(a-0).

f'(x)=1-3/(x^2)

Derivative of f at point a must equal the slope of the tangent line, i.e. we must have
f'(a)=1-3/(a^2)=(a+3/a-4)/(a-0)=m

Solving I get a=3/2. However, looking at the graph of f(x), it seems there should be two points where the tangent line passes through 4, the other one being on the part of the f(x) where x<0. Where did I go wrong?

I don't think you went wrong anywhere. Looking at the graph I don't see any points x<0 where the tangent line will go through (0,4).
 
If a is negative, the y intercepts of the tangent lines are also negative.

ehild
 

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