Question

Question #e0971

Answer 1

`" "e^x\approxx+1` for `x\approx0`
`\thereforee^-0.015\approx-0.015+1=0.985`

Explanation:

We know that `d/dx(e^x)=e^x` and `e^0=1`.

Then, we can find that the tangent line to `e^x` at `x=0` has a slope of `e^0=1` and has a y-intercept at `(0,e^0)=(0,1)`. Thus, the equation of the tangent line is `y=x+1`.

Graph the tangent line and zoom in:
graph{(y-x-1)(y-e^x)=0 [-0.5,0.5,-0.5,1.5]}

It seems that the line `y=e^x` and `y=x+1` are quite close in near `x=0`. In other words, `e^x\approxx+1` when `x\approx0`.

So, `e^-0.015\approx-0.015+1=0.985`, which is quite close to the actual value `e^-0.015=0.9851` (to four decimal places).