Question

What is the equation of the line tangent to `f(x)=1/sqrt(e^x-3x)` at `x=0`?

Answer 1

The equation is `y = x + 1`.

Explanation:

At `x= 0`, the function has y-value

`f(0) = 1/sqrt(e^0 - 3(0)) = 1/sqrt(1) = 1`

We now rewrite the function as

`f(x) = (e^x- 3x)^(-1/2)`

We can find this derivative using the chain rule.

`f'(x) = -1/2u^(-3/2) * (e^x - 3) = -(e^x - 3)/(2(e^x - 3x)^(3/2)`

The slope of the tangent is hence

`f'(0) = -(e^0 - 3)/(2(e^0 - 3(0))^(3/2))`

`f'(0) = - (-2)/(2)`

`f'(0) = 2/2`

`f'(0) = 1`

Now write the equation of the tangent.

`y -y_1 = m(x - x_1)`

`y - 1 = 1(x - 0)`

`y = x + 1`

Hopefully this helps!