SOLUTION: Consider the function f(x) = 1/x I cannot use derivatives, my teacher requires using limits or other methods to solve. (a). Find the slope of the L tangent to the graph of (3

Algebra ->  Equations -> SOLUTION: Consider the function f(x) = 1/x I cannot use derivatives, my teacher requires using limits or other methods to solve. (a). Find the slope of the L tangent to the graph of (3      Log On


   

Question 988918: Consider the function f(x) = 1/x
I cannot use derivatives, my teacher requires using limits or other methods to solve.
(a). Find the slope of the L tangent to the graph of (3,f(3))
(b). Find an equation of the line L
(c). Find the x-intercept and y-intercept of L
Please explain how these are figured out. Very confused.
Also, what exactly is 3,f(3) are these just arb pts?
Thank you
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The point (3,f(3)) is the same as (3,1/3) since f(3) = 1/3


Plug x = 3 into f(x) = 1/x to get f(3) = 1/3


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Part a)


f%28x%29+=+1%2Fx


f%28x%2Bh%29+=+1%2F%28x%2Bh%29


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Let's simplify the difference quotient


%28f%28x%2Bh%29-f%28x%29%29%2Fh


%281%2F%28x%2Bh%29-1%2Fx%29%2Fh


%28x%2F%28x%28x%2Bh%29%29-1%2Fx%29%2Fh


%28x%2F%28x%28x%2Bh%29%29-%28x%2Bh%29%2F%28x%28x%2Bh%29%29%29%2Fh


%28%28x-%28x%2Bh%29%29%2F%28x%28x%2Bh%29%29%29%2Fh


%28x-%28x%2Bh%29%29%2F%28xh%28x%2Bh%29%29


%28x-x-h%29%2F%28xh%28x%2Bh%29%29


%28-h%29%2F%28xh%28x%2Bh%29%29


-1%2F%28x%28x%2Bh%29%29


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Now apply the limit.


As h approaches 0, -1%2F%28x%28x%2Bh%29%29 will approach -1%2F%28x%28x%2B0%29%29+=+-1%2F%28x%2Ax%29+=+-1%2F%28x%5E2%29


So the derivative function is -1%2F%28x%5E2%29


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Plug x = 3 into the derivative function to get -1%2F%28x%5E2%29=-1%2F%283%5E2%29=-1%2F9


Therefore, the slope of line L is m=-1%2F9


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Part b)


The slope of line L is -1%2F9 (found earlier). The tangent line L goes through (3,1/3) so x = 3 and y = 1/3.


Plug in m+=+-1%2F9, x=3 and y=1%2F3 into y=mx%2Bb. Then solve for b.


y=mx%2Bb


1%2F3=%28-1%2F9%29%2A%283%29%2Bb


1%2F3=%28-1%2F9%29%2A%283%2F1%29%2Bb


1%2F3=-3%2F9%2Bb


1%2F3=-1%2F3%2Bb


1%2F3%2B1%2F3=b


2%2F3=b


b=2%2F3


The equation of the tangent line L is y+=+%28-1%2F9%29x%2B2%2F3


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Part c)


The y-intercept is 2/3 or the point (0,2/3). This was found in part b) above.


The x-intercept is found by plugging in y = 0 and solving for x


y+=+%28-1%2F9%29x%2B2%2F3


0+=+%28-1%2F9%29x%2B2%2F3


0-2%2F3+=+%28-1%2F9%29x%2B2%2F3-2%2F3


-2%2F3+=+%28-1%2F9%29x


%28-9%2F1%29%2A%28-2%2F3%29+=+%28-9%2F1%29%2A%28-1%2F9%29x


6+=+x


x=6


The x-intercept is 6, which is the same as the point (6,0)


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Graph


f%28x%29+=+1%2Fx is in green


y+=+%28-1%2F9%29x%2B2%2F3 is in blue (tangent line at (3,1/3))


+graph%28+500%2C+500%2C+-5%2C+5%2C+-5%2C+5%2C+0%2C1%2Fx%2C%28-1%2F9%29%2Ax%2B2%2F3%29+


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If you need more help, or if you have any questions about the problem, feel free to email me at
jim_thompson5910@hotmail.com