The given curve is --- y = x^5 + x^3 - 2x
First derivative to this curve is y' = 5 x^4 + 3 x^2 - 2
=> Slope (m) = 5 x^4 + 3 x^2 - 2
For the minimum value of m we calculate dm/dx and put it = 0
=> d( 5 x^4 + 3 x^2 - 2 ) / dx = 0
=> 20 x^2 + 6 x = 0
=> x ( 20 x + 6 ) = 0
Turning values of x are 0 and - 6 / 20 = ( - 3 / 10 )
At x = 0 , m = - 2
and at x = - 3/10
m = 5 x^4 + 3 x^2 - 2
=> m = 5 ( - 3 / 10 )^4 + 3 ( - 3 / 10 )^2 - 2 = - 1.68
Hence at turning points, the slope is minimum at x = 0 and is equal to = - 2
MINIMUM VALUE OF THE SLOPE = - 2