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

1 Answer
Nov 11, 2015

#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#.