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anonymous
 4 years ago
Find the equation of the tangent line to the graph of the function y=f(x)=x^2(x2)^3 at the point with x=1.
anonymous
 4 years ago
Find the equation of the tangent line to the graph of the function y=f(x)=x^2(x2)^3 at the point with x=1.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is \[f (x)=x ^{2}\times(x2)^{3}\] ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0derive the function and evaluate in 1 and you'll have the pendient, replace 1 in the first function and you have the point, so you have the pendient and one point, you can get the equation (ya)=m(xb), sorry for my english I speak spanish XD

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do I start to evaluate the function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0step by step ...first: replace 1 in the function, what is the result?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok you have the point (1, 1) , remember the equation of a rect you need a point and the pendient, you have a point, now you need the pendient in x=1, the pendient in x=1 is the derivate of the function evaluate in 1, .. can you derive the function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know past here...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok the derivate of the function is : \[2x(x3)^{3} + x ^{2}(3(x2)^{2})\] if you evaluate in 1 the result is 13, I think, so you have the pendient m= 13 and the point (a,b)= (1, 1) .. the rect is (ya)=m(xb) and replace the values

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i have a mistake is 2x(x2)^3 but is the same idea
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