Question

At what point, the slope of the curve y= x^3 + 3x^2+ 9x - 27 is maximum? also find the maximum slope

Answer 1

Step 1

f(r)=−x3+3x2+9x−27

Differentiating w.r.t x we get

f′(x)=−3x2+6x+9

If f′(x)=0

⇒−3x2+6x+9=0

3(−x2+2x+3)=0

factorising this we get,

f′(x)=(x+3)(x−1)

when x=−3

f(x)=−(3)3+3(3)2+9(3)−27

=−27+27+27−27

=0

when x=+1

f(1)=−(+1)3+3(+1)2+9(+1)−27

=−16

Hence the point is (1,−16)

The maximum slope =

f′(1)=−3(1)+6(1)+9

=12

Answer 2

$\bf\huge\ Question:-$

At what point, the slope of the curve y= x^3 + 3x^2+ 9x - 27 is maximum? also find the maximum slope

$\bf\huge\ Answer:-$

your answer is 12