What is the equation of the line tangent to f(x)=x2−xex−3x at x=0?

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carlosegundoacyg · Accepted Answer

The y coordinate of the function at x=0 is:

y = f (0) = 0^{2} - 0 \sqrt{e^{2} - 3 (0)}

y=0
Therefore, the line must contain the point (0,0)
Compute the first derivative:

f^{'} (x) = 2 x - \sqrt{e^{x} - 3 x} - \frac{x (e^{x} - 3)}{2 \sqrt{e^{x} - 3 x}}

The slope, m, of the line is the first derivative of the function evaluated at the x coordinate, x=0. Please of observe that, if you substitute 0 for all of the xs in the above equation, the middle term becomes -1 and the other terms are 0:
m=-1
We have enough information to use the point slope form of the equation of a line to write the equation of the line:
y=-1(x-0)+0
This simplifies:
y=-x

What is the equation of the line tangent

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