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07302007  #1 
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calculus question?
the tangent line approximation to the function f(x)= 1/x^2 at the point (1,1) is? (can you explain how you got it?)

07302007  #2 
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to arrive at ttangent to a curve at a specific point, differentiate the equation and plug in the point to find the slope of tangent at that point. then form the tangent equationdf(x)/dx = 2/x^3at (1,1), df(1)/dx = 2so tangent at (1,1) is y1 = 2(x1)y + 2x = 3

07302007  #3 
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the answer is x+y=2y=f(x)=1/x^2slope =dy/dx=f'(x)=1/x^3slope at (1,1)=1/1^3=1eqn of line witha slope of 1 and passing through (1,1) is y1=1(x1)y1=x+1 =>x+y=2.

07302007  #4 
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1. take the dirivative of the function using the quotient rule dy/dx= 2x/x^42. use the x point and plug it into the dy/dx equation to find the slope at that point dy/dx= 2(1)/1^4= 23. then you have that equation of a line, y=mx+b and you can plug in what you know to solve for b 1= 2(1)+b b=3 (the x and y value come from the point given in the problem and the m value is the derivative evaluated at that point)4. so the equation on the tangent line is y= 2x+3

07302007  #5 
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derivative of the given function dy/dx= 2/x^3 (nx^[n1])so the slope's = 2/x^3when x=1 (from point 1,1)slope = 2/1 = 2so the equation of a line as we know y=mx+cwhen passing thru one pointyy1=m(xx1)y1=1;x1=1 (from the point given)slope m we have found out = 2so the equation of the line : y1= 2(x1) => y1= 2x+2 =>y+2x= 2 (bringing x to left,and 2 to right)or 2x+y = 2hope its fully clear now

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