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

Answer 1

Christopher Allers

#y=143x-381#

At every point, the derivative's value is equal to the gradient of the tangent.

#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.

Changing the point located in we obtain

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

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

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Answer 2

Charles Arocha

The equation of the tangent line of f(x) = 3x^3+e^(-3x)-x at x=4 is y = 192x - 257.

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Answer from HIX Tutor

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