Question

What is the equation of the line tangent to `f(x)=x^2-xsqrt(e^x-3x)` at `x=0`?

Answer 1

`y = -x`

Explanation:

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

`y = f(0) = 0^2- 0sqrt(e^2-3(0))`

`y = 0`

Therefore, the line must contain the point `(0,0)`

Compute the first derivative:

`f'(x) = 2x - sqrt(e^x-3x)-(x(e^x-3))/(2sqrt(e^x-3x))`

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`