Finding largest and smallvest value

[K_Swiss]

Find the largest and smallest values of the given function over the prescribed closed, bounded interval.

\(\displaystyle f(x) = \frac{ln(x + 1)}{x + 1}\) for 0 ? x ? 2

\(\displaystyle f'(x) = \frac{\frac{1}{x + 1}(1)(x + 1) - (1)(ln(x + 1))}{(x + 1)^{2}}\)

\(\displaystyle f'(x) = \frac{\frac{x + 1}{x + 1} - ln(x + 1)}{(x + 1)^{2}}\)

\(\displaystyle f'(x) = \frac {1 - ln(x + 1)}{(x + 1)^{2}}\)

\(\displaystyle 0 = \frac {1 - ln(x + 1)}{(x + 1)^{2}}\)

\(\displaystyle 0 = 1 - ln(x + 1)\)
\(\displaystyle ln(x + 1) = 1\)
...don't know what to do now.

\(\displaystyle (x + 1)^{2} = 0\)
\(\displaystyle (x + 1)(x - 1) = 0\)
\(\displaystyle x = 1\), \(\displaystyle x = -1\)

\(\displaystyle f(0) = 0\)

\(\displaystyle f(2) = 0.3662\)

\(\displaystyle f(1) = 0.34657\)

\(\displaystyle f(-1) = error\)

 

[K_Swiss]

...textbook answers are
\(\displaystyle \frac{1}{e}\), \(\displaystyle 0\)

I don't know how to get to those answers.

 

[galactus]

You're on the right track. If you differentiate, as you done, you get

\(\displaystyle ln(x+1)=1\)

Solving, gives \(\displaystyle x=e^{1}-1\)

Plug this into your original to get the max value.

Look at the graph if nothing else. See where the 0 comes from?.