Question

What is the equation of the tangent line of `f(x) = 3x^3+e^(-3x)-x` at `x=4`?

Answer 1

`y=143x-381`

Explanation:

The gradient of the tangent at a certain point equals the value of the derivative at that point.

`therefore d/dx (3x^3+e^(-3x)-x)=9x^2-3e^(-3x)-1`

Hence `f'(4)=9*4^2-3e^(-3*4)-1=143-3/e^12=~~143`

But the actual function `f(4)=3*4^3+e^(-3*4)-1=191`

So thus the point `(4, 191)` is the point of contact of the function f with the tangent at that point.

Since the tangent is a straight line, it must satisfy the linear equation `y=mx+c` where m is the gradient and c is the y-intercept.

Substituting the point found in we get

`191=143*4+c`, from which `=>c=-381.`

Therefore the tangent to the function has equation `y=143x-381`.