Maximum slope of the curve $y = - x^{3} + 3 x^{2} + 9 x - 27$ is :

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $12$
$y = - x^{3} + 3 x^{2} + 9 x - 27$
$\frac{d y}{d x} = - 3 x^{2} + 6 x + 9$ = Slope of the curve

$\frac{d^{2} y}{d x^{2}} = - 6 x + 6 = - 6 (x - 1)$

Now, $\frac{d^{3} y}{d x^{3}} = - 6 < 0$

So, the slope is maximum when $x = 1$ .
Therefore,
$(\frac{d y}{d x})_{x = 1} = - {3.1}^{2} + 6.1 + 9 = 12$

Suggest Corrections

Similar questions

y = - x^{3} + 3 x^{2} + 9 x - 27

y = - x^{3} + 3 x^{2} + 9 x - 27

At what point is the slope of the curve $y = - x^{3} + 3 x^{2} + 9 x - 27$ maximum? Also, find the maximum slope.

Join BYJU'S Learning Program