Find the equation of the curve y = F (x) which passes through the point and is indicated as...?

Find the equation of the curve y = F (x) which passes through the point and is indicated as the slope of the tangent at any point x of this curve is given the function: x^3 - x^2 - x - 1 (-1, 1)

If you mean the slope of the tangent line (derivative) is x^3-x^2-x-1,

Then integrate to find F(x)

F(x)=( x^4)/4 -(x^3)/3 - (x^2)/2 -x + C

Since F(-1)= 1 as given by (-1,1) Then plug in -1 for x to find C

1= 1/4 +1/3 -1/2 +1 +C

0= 1/12 +C

C= -1/12

F(x) = (1/4)x^4 -(1/3)x^3 -(1/2)x^2 -x-1/12

Hoping this helps!

find an equation of the tangent to the curve at the given point y=4-x^2 ,(-1 ,3)

f(x) = 8/?(x - 2) f'(x) = -4/(x - 2)?(x - 2) f(6) = 4 f'(6) = -4/(6 - 2)?(6 - 2) = -a million/2 The equation of the tangent line at (6,4) is y - 4 = (-a million/2)(x - 6) y = (-a million/2)x + 7