Question

What is the equation of the line tangent to `f(x)= x-e^x` at `x=4`?

Answer 1

`y-(4-e^4)=(1-e^4)(x-4)`

Explanation:

Given that `f(x)=x-e^x`

Find that `f'(x)=1-e^x`

`f(4)=4-e^4`
`f'(4)=1-e^4`

Thus, the tangent line passes through the point `(4,4-e^4)` and has a slope of `1-e^4`.

Write this in point-slope form:

`y-(4-e^4)=(1-e^4)(x-4)`