At what point, the slope of the curve y = –x³+3x²+9x–27 is maximum? Also find the maximum slope.

Given: y = –x³+3x²+9x–27

To find: at what point the slope of the curve is maximum, and to also find the maximum value of the slope

Explanation: given y = –x³+3x²+9x–27

By finding the first derivative of the given equation of the curve, we get the slope of the curve.

So slope of the curve is

Now applying the derivative, we get

Now to find the critical point we need to find the derivative of the slope, so

Applying the derivative, we get

Now critical point is found by equating the second derivative to 0, i.e.,

⇒ -6x+6 = 0

⇒ 6x = 6

⇒ x = 1…….(i)

Now we will find the third derivative of the given curve,

i.e.,

Applying the derivative, we get

As the third derivative is less than 0, so the maximum slope of the given curve is at x = 1.

To find the value of the maximum slope we will find first derivative at x = 1, i.e.,

Hence the slope of the curve is maximum at x = 1, and the maximum value of the slope is 12.