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Old 07-30-2007   #1
Sarah K
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Default 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?)
Old 07-30-2007   #2
Vipin A
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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 y-1 = -2(x-1)y + 2x = 3
Old 07-30-2007   #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 y-1=-1(x-1)y-1=-x+1 =>x+y=2.
Old 07-30-2007   #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
Old 07-30-2007   #5
sonny torreli
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derivative of the given function dy/dx= -2/x^3 (nx^[n-1])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 pointy-y1=m(x-x1)y1=1;x1=1 (from the point given)slope m we have found out = -2so the equation of the line : y-1= -2(x-1) => y-1= -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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