Question

Find the Absolute max and min for
fons = ln(1+x) over the interval [0, 2]​

Answer 1

Answer:

Let's find, for example, the absolute extrema of h(x)=2x^3+3x^2-12xh(x)=2x3+3x2−12xh, left parenthesis, x, right parenthesis, equals, 2, x, cubed, plus, 3, x, squared, minus, 12, x over the interval -3\leq x\leq 3−3≤x≤3minus, 3, is less than or equal to, x, is less than or equal to, 3.

h'(x)=6(x+2)(x-1)h′(x)=6(x+2)(x−1)h, prime, left parenthesis, x, right parenthesis, equals, 6, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, so our critical points are x=-2x=−2x, equals, minus, 2 and x=1x=1x, equals, 1. They divide the closed interval -3\leq x\leq 3−3≤x≤3minus, 3, is less than or equal to, x, is less than or equal to, 3 into three parts:



\llap{-}3-3\llap{-}2-2\llap{-}1-100112233\llap{-}3 <x< \quad \llap{-}2-3<x<-2\llap{-}2<x<1-2<x<11<x<31<x<3

Intervalxxx-valueh'(x)h′(x)h, prime, left parenthesis, x, right parenthesisVerdict-3<x<-2−3<x<−2minus, 3, is less than, x, is less than, minus, 2x=-\dfrac52x=−25x, equals, minus, start fraction, 5, divided by, 2, end fractionh'\left(-\dfrac52\right)=\dfrac{21}{2}>0h′(−25)=221>0h, prime, left parenthesis, minus, start fraction, 5, divided by, 2, end fraction, right parenthesis, equals, start fraction, 21, divided by, 2, end fraction, is greater than, 0hhh is increasing \nearrow↗\nearrow