Examine the function f(x)=x+(1/x) a) Find the points on the curve at which the tangent lines pass through the point (1,1) b) Prove that no tangent line to the curve passes through the origin c) Prove that there is one tangent line passing through the point (A,A), A (does not equal)0, and it occurs at the point with x-coordinate (A/2).
\[f(x)=x+\frac{1}{x}\] \[f'(x)=1-\frac{1}{x^2}\]
part B set \[1-\frac{1}{x^2}=0\] and try to solve for x
oh yey its you again
I assume its 1
hold on maybe i made a mistake
yeah these questions are for the lose
ok first one
point on the curve is \[(x,x+\frac{1}{x})\] and if it goes through (1,1) the slope is \[\frac{x+\frac{1}{x}-1}{x-1}\] and you know this has to equal the derivative so set \[\frac{x+\frac{1}{x}-1}{x-1}=1-\frac{1}{x^2}\]and solve
actually easier than it looks get \[x=\frac{1}{2}\]
I just did it. :) and got it
Aahh can't believe I found this! Thanks so much in advance! The answer to (a) is supposed to be (1/2, 5/2), but I can't figure out how to get there.
UF?
Yep lol (:
you got 1/2 yes?
yep.
now repeat w/ (0,0) insted of (1,1) and see the trouble
is it because you can't do (0, 0+1/0)?
write\[\frac{x+\frac{1}{x}}{x}=1-\frac{1}{x^2}\] and you will not be able to solve
left hand side is slope at (0,0), right hand side is derivative. you cannot solve this
ok right.
and for part c do I plug in A where the x's were, so (A+(1/A))/A=1-(1/A^2)
?
no for part c you plug in A in the formula for the slope, just like we plugged in (1,1)
\[\frac{x+\frac{1}{x}-A}{x-A}=1-\frac{1}{x^2}\] solve for x
ok I did that and got A/2
how do I prove that that's the only tangent line.
and the point is (a/2,1)
you did it when you solved and got a/2 ! you are done
:) now what's the best way to sketch this curve and illustrate those ideas?
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