Given that F(x)=x^3 + 2; find the equation of the tangent line at the point (-1,1)?

Question

Just a basic Calculus problem :D

Answer 1

f `(x) = 3 x ²
m = f `(- 1) = 3
y - 1 = 3( x + 1)
y = 3x + 4

Answer 2

Create a table of result for the equation
Plot the curve
Draw a line at a tangent at (-1,1)
Take two points on the line
Use the line equation

equation of Tangent to curve at (-1,1)
y = 2.85x +4

Answer 3

f(x)=x^3 + 2

d / df(x) = 3x^2

When x = -1,

d / df(x) = 3 (gradient of tangent)

Using point (-1,1)

y=mx+c
1=3(-1) + c
4=c

Therefore the equation of the line is
y=3x+4

Answer 4

Take the derivative of this f and you get F'(x)=3x^2
In x=-1, F'(-1)=3=slope of tangent line.
y-y1/x-x1=3
y-1/x+1=3
y-1=3x+3
y=3x+4

Answer 5

It's F(y) not F(x)