At what points, the slope of the curve `y=-x^3+3x^2+9x-27` at point `(x , y)` is given by maximum slope.
At what points, the slope of the curve `y=-x^3+3x^2+9x-27` at point `(x , y)` is given by maximum slope.
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We have, `y=-x^(3)+3x^(2)+9x-27`
`therefore (dy)/(dx)=-3x^(2)+6x+9`= slope of tangent to the curve
Now, `(d^(2)y)/(dx^(2))= -6x+6`
For `d/(dx)(dy)/(dx)=0`,
`-6x + 6=0`
`rArr x=-6/-6=1`
`therefore d/(dx) (d^(2)y)/(dx^(2))=-6 lt 0`
So, the slope of tangent to the curve is maximum, when x=1.
For x=1, `(dy)/(dx)_(x=1) = -3.1^(2)+6.1+9=12,`
Which is maximum slope.
Also, for `x=1, y=-1^(2)+3.1^(2)+9.1-27`
`=-1 + 3+9-27`
`=-16`
So, the required point is `(1,-16)`.
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