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mariina.1319

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If -2 is slope of normal, then -1/(-2) or 1/2 is the slope of the tangent, which is the derivative at some value x0. Therefore, write an equation so you can solve for x0. Then plug that point (x0, arctan(x0-1)) into the equation y = -2x + c to find y-intercept of the equation of the normal.

For your other question, reproduced below

The normal to the curve y=(k/x)+lnx^2, for x≠0, at the point when x=2, has the equation 3x+2y=b. Find the exact value of k.

Here the normal is given in standard form of a line, you need it in another form to be read off the slope. Similarly as previous question, find the negative reciprocal of the normal slope, which would be the tangent slope or the derivative. Equate this derivative with the derivative function y' and plug in x = 2. Solve for k. 

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13 hours ago, kw0573 said:

If -2 is slope of normal, then -1/(-2) or 1/2 is the slope of the tangent, which is the derivative at some value x0. Therefore, write an equation so you can solve for x0. Then plug that point (x0, arctan(x0-1)) into the equation y = -2x + c to find y-intercept of the equation of the normal.

For your other question, reproduced below

The normal to the curve y=(k/x)+lnx^2, for x≠0, at the point when x=2, has the equation 3x+2y=b. Find the exact value of k.

Here the normal is given in standard form of a line, you need it in another form to be read off the slope. Similarly as previous question, find the negative reciprocal of the normal slope, which would be the tangent slope or the derivative. Equate this derivative with the derivative function y' and plug in x = 2. Solve for k. 

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Can you give a complete solution and explanation of the problem:  "The normal to the curve y=(k/x)+lnx^2, for x≠0, at the point when x=2, has the equation 3x+2y=b. Find the exact value of k." PLEASE!!!! @kw0573I really need it!      Thank you very very much :) 

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