Question

What is the equation of the line tangent to `f(x)=2x^3 - 5x` at `x=-1`?

Answer 1

Equation of line tangent is `x-y+4=0`

Explanation:

As at `x=-1`, `f(x)=2(-1)^3-5(-1)=-2+5=3`,

we are seeking tangent at the point `(-1,3)`

Slope of tangent is given by the value of first derivative at the poiint we seek tangent and

as `f(x)=2x^3-5`, `f'(x)=(df)/(dx)=2xx3x^2-5=6x^2-5`

and as `f'(-1)=6(-1)^2-5=1`

Hence, equation of tangent is

`y-3=1(x-(-1))` or `y-3=x+1` or `x-y+4=0`

graph{(2x^3-5x-y)(x-y+4)=0 [-10.54, 9.46, -4.12, 5.88]}