Calculus Help!!!?

Question

A curve is given by the equation y=(3x^​2+1)e^​((1/3)x^​3-3x^​2+1) 

a. Compute the slope of the tangent to the curve at any point 

b. Compute the x-coordinates of all points at which the tangent to the curve is 
horizontal. 

Answer 1

(a)

Curve of the Equation is

To find the slope of the tangent, We differentiate y with respect to x.

Therefore the Slope of the Tangent at any Point is

.

Answer 2

(b)

Curve of the Equation is

To find the slope of the tangent, We differentiate y with respect to x.

The Slope of the Tangent at any Point is .

Tangent to the Curve is Horizantal means y' = 0

                (Since Exponents cannot be equated to zero)

x²(3x²-18x+1) = 0

If x² = 0 then x = 0,0

If 3x²-18x+1 = 0 then

Roots are 5.94 and 0.560

Therefore the Values of x are 0, 0, 0.560 and 5.94.