Slop of Curve

Question

 please share your solution 

The maximum slope of the curve
-x³+6x²+66x+666

Comments

maximum value of slope, $m_{max}=78$ at $x=2$

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solution???

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$f(x)=-x^3+6x^2+66x+666$

equation of slope at point $x$, $m=f'(x)=-3x^2+12x+66$

to find max/min value of slope, put $\frac{dm}{dx}=0$

$\frac{dm}{dx}=-6x+12=0 , x=2$

now, $\frac{d^2m}{dx^2}=-6<0$, therefore at $x=2$, slope m is maximum.

therefore max slope at $x=2$ is $m_{max}=78$

Answer 1

$f(x)=-x^3+6x^2+66x+666$

equation of slope at point $x$, $m=f'(x)=-3x^2+12x+66$

to find max/min value of slope, put $\frac{dm}{dx}=0$

$\frac{dm}{dx}=-6x+12=0 , x=2$

now, $\frac{d^2m}{dx^2}=-6<0$, therefore at $x=2$, slope m is maximum.

therefore max slope at $x=2$ is $m_{max}=78$