You need to differentiate the function and then you should solve the equation f'(x) = 0 to find the absolute extrema of function such that:

`f'(x) = -6e^(-6x) + 5e^(-5x)`

You need to solve the equation f'(x) = 0 such that:

`-6e^(-6x) + 5e^(-5x) = 0`

You need to factor out `e^(-5x)` such that:

`-e^(-5x)(6e^(-6x+5x) - 5) = 0`

Since `e^(-5x) != 0 =gt 6e^(-6x+5x) - 5 = 0`

`6e^(-x) = 5 =gt e^(-x) = 5/6`

`ln e^(-x) = ln (5/6) =gt -x =ln (5/6) =gt x = ln(6/5)`

You need to substitute 1 for x in f'(x) such that:

`f'(1) = -6e^(-6) + 5e^(-5) =gt f'(1) = 5/(e^5) - 6/(e^6)`

`f'(1) = (5e-6)/(e^6) gt 0`

You need to substitute 3 for x in f'(x) such that:

`f'(3) = -6e^(-18) + 5e^(-15)`

`f'(3) = 5/(e^15) - 6/(e^18) gt 0`

**Hence, the function is increasing over interval [1,3] and the function reaches its absolute extrema at `x = ln(6/5).` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now