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Find the slope of the tangent to the curve y=3+4x^2-x^3 at the point where x=a.  If anyone knows how to solve this problem and can give step-by-step instructions, that would be great.  Thanks.

gibsonj338  Apr 18, 2018
 #1
avatar+88899 
+1

y  =  3 + 4x^2  -  x^3

 

The derivative of this is the slope at any point, a

 

So

 

y'  =  0 + 8x  - 3x^2        simplify

 

y'  = -3x^2 + 8x

 

so...the slpoe at x  = a   is

 

-3(a)^2   + 8a

 

 

cool cool cool

CPhill  Apr 18, 2018

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