435 views
3 votes
3 votes

 please share your solution 

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

1 Answer

Best answer
1 votes
1 votes

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

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 $m_{max}=78$

selected by

Related questions

1 votes
1 votes
0 answers
1
Tuhin Dutta asked Jan 25, 2018
596 views
x00.30.60.91.21.51.82.12.4f(x)00.090.360.811.442.253.244.415.76The value of the below integral computed using the continuous at x = 3?$$\int_{0}^{3} f(x) dx$$a) 8.983b) 9...
1 votes
1 votes
1 answer
2
Shubhanshu asked Sep 8, 2017
826 views
Let f(x)=x−(1/2) and A denote the area of region bounded by f(x) and the x-axis, when x varies from -1 to 1.A is nonzero and finite??
1 votes
1 votes
2 answers
3
sh!va asked Mar 10, 2017
976 views
The area bounded by the curves $y^2$ = 9x, x - y + 2 = 0 is given bya) 1b) 1/2C) 3/2d) 5/4
0 votes
0 votes
1 answer
4
sh!va asked Mar 8, 2017
1,736 views
Area bounded by the parabola 2y= $x^2$ and the line x = y-4 is equal to(a) 4.5(b) 9(c) 18(d) 36