What is the equation of the tangent line of f(x)=3x3+e−3x−x at x=4?

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pautmndu · Accepted Answer

The derivative value at a given place is equal to the gradient of the tangent there.
$∴ \frac{d}{d x} (3 x^{3} + e^{- 3 x} - x) = 9 x^{2} - 3 e^{- 3 x} - 1$
Hence $f^{'} (4) = 9 \cdot 4^{2} - 3 e^{- 3 \cdot 4} - 1 = 143 - \frac{3}{e^{12}} =\approx 143$
But the actual function $f (4) = 3 \cdot 4^{3} + e^{- 3 \cdot 4} - 1 = 191$
So thus the point (4, 191) is the point of contact of the function f with the tangent at that point.
The linear equation y=mx+c, where m is the gradient and c is the y-intercept, must be satisfied by the tangent because it is a straight line.
By changing the position in we obtain
191=143*4+c, from which $\Rightarrow c = - 381$ .
Hence the tangent to the function has equation y=143x-381.

What is the equation of the tangent line

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