Finding largest and smallvest value

K_Swiss

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Feb 8, 2008
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Find the largest and smallest values of the given function over the prescribed closed, bounded interval.

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

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

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

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

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

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

(x+1)2=0\displaystyle (x + 1)^{2} = 0
(x+1)(x1)=0\displaystyle (x + 1)(x - 1) = 0
x=1\displaystyle x = 1, x=1\displaystyle x = -1

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

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

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

f(1)=error\displaystyle f(-1) = error
 
...textbook answers are
1e\displaystyle \frac{1}{e}, 0\displaystyle 0

I don't know how to get to those answers.
 
You're on the right track. If you differentiate, as you done, you get

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

Solving, gives x=e11\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?.
 
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