Question

How do you find the slope of a tangent line to the graph of the function `f(x) = 2e^(-3x)` at (0, 2)?

Answer 1

`f'(0)=-6`

Explanation:

`f(x)=2e^(-3x)`
Use the exponent rule to differentiate `f(x)`:
`f'(x)=(-3)(2)(e^(-3x))`
`f'(x)=-6e^(-3x)`

`f'(0)=-6e^[-3*0]=-6`