Question 988918
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(x) = 1/x}}}



{{{f(x+h) = 1/(x+h)}}}



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Let's simplify the difference quotient



{{{(f(x+h)-f(x))/h}}}



{{{(1/(x+h)-1/x)/h}}}



{{{(x/(x(x+h))-1/x)/h}}}



{{{(x/(x(x+h))-(x+h)/(x(x+h)))/h}}}



{{{((x-(x+h))/(x(x+h)))/h}}}



{{{(x-(x+h))/(xh(x+h))}}}



{{{(x-x-h)/(xh(x+h))}}}



{{{(-h)/(xh(x+h))}}}



{{{-1/(x(x+h))}}}



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Now apply the limit.



As h approaches 0, {{{-1/(x(x+h))}}} will approach {{{-1/(x(x+0)) = -1/(x*x) = -1/(x^2)}}}



So the derivative function is {{{-1/(x^2)}}}



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Plug x = 3 into the derivative function to get {{{-1/(x^2)=-1/(3^2)=-1/9}}}



Therefore, the slope of line L is {{{m=-1/9}}}



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Part b)



The slope of line L is {{{-1/9}}} (found earlier). The tangent line L goes through (3,1/3) so x = 3 and y = 1/3.



Plug in {{{m = -1/9}}}, {{{x=3}}} and {{{y=1/3}}} into {{{y=mx+b}}}. Then solve for b.



{{{y=mx+b}}}



{{{1/3=(-1/9)*(3)+b}}}



{{{1/3=(-1/9)*(3/1)+b}}}



{{{1/3=-3/9+b}}}



{{{1/3=-1/3+b}}}



{{{1/3+1/3=b}}}



{{{2/3=b}}}



{{{b=2/3}}}



The equation of the tangent line L is {{{y = (-1/9)x+2/3}}}



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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 = (-1/9)x+2/3}}}



{{{0 = (-1/9)x+2/3}}}



{{{0-2/3 = (-1/9)x+2/3-2/3}}}



{{{-2/3 = (-1/9)x}}}



{{{(-9/1)*(-2/3) = (-9/1)*(-1/9)x}}}



{{{6 = x}}}



{{{x=6}}}



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



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Graph



{{{f(x) = 1/x}}} is in green



{{{y = (-1/9)x+2/3}}} is in blue (tangent line at (3,1/3))



{{{ graph( 500, 500, -5, 5, -5, 5, 0,1/x,(-1/9)*x+2/3) }}}



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If you need more help, or if you have any questions about the problem, feel free to email me at 
<font color="blue"><code>jim_thompson5910@hotmail.com</code></font>