Question

The maximum value of slope of the curve y=-x³ + 3x²+12x - 5 is

Answer 1

The maximum value of slope will be - 0.16

Step-by-step explanation:

Given as :

The curve equation is

 y = -  x³  + 3  x² + 12  x  - 5

For maximum value of slope of the curve

we make slope =  $\dfrac{dy}{dx}$   should be   0

The equation of a slope at evry popint , or let it be a fuction which tells slope at any point .

So,  Differentiate y with respect to x

Or,  $\dfrac{d(-x^{3} + 3x^{2} +12 x -5)}{dx}$   = 0

Or,   $\dfrac{d(-x^{3} )}{dx}$ + 3 $\dfrac{dx^{2} }{dx}$  + 12 $\dfrac{dx}{dx}$ + $\dfrac{d(-5)}{dx}$  = 0

Or,  - 3 x²  + 3 ( 2 x ) + 12 ( 1 ) + 0 = 0

Or,   - 3 x²  + 6 x  + 12  + 0 = 0

Taking - 3 as common

i.e     - 3 ( x²  - 2 x -  4 )  = 0

Or,    ( x²  - 2 x  - 4 )  = 0

Solving this quadratic equation as form of a x²  + b x  + c = 0

       x = $\dfrac{2\pm \sqrt{4-4\times 1\times (-4)}}{2\times 1}$

i.e     x = $\dfrac{2\pm \sqrt{20}}{2}$

Or,     x = 3.2  ,  - 1.2

Again

        $\dfrac{d{2}y }{dx^{2} }$   = $\dfrac{d(x^{2} -2 x -4)}{dx}$

                 = $\dfrac{dx^{2} }{dx}$  - 2 $\dfrac{dx}{dx}$ - $\dfrac{d(4)}{dx}$

                 = 2 x - 2

 ∵     $\dfrac{d{2}y }{dx^{2} }$ = 0

So,     2 x - 2  = 0

Or,      2 x = 2

∴            x = $\dfrac{2}{2}$

i.e           x = 1

So, The maximum value of slope will be at x = 3.2

So, slope = $\dfrac{dy}{dx}$  is maximum at x = 3.2

i.e           slope  =  (3.2)²  - 2 (3.2)  - 4

                         = 10.24 - 6.4 - 4

                          = - 0.16

So, The maximum value of slope = - 0.16

Hence, The maximum value of slope will be - 0.16  Answer

Answer 2

Answer:

Step-by-step explanation: