Find the maximum value of g(x) = 12 - ln(x^2 + 1 ) on the closed interval [-1,5]

Answer 1

Hi Danielle,

this question has a beautiful point.

g(x) = 12 - ln(x^2 + 1 )  [-1,5]

First, find derivative and x of local min or max

g'(x) = -2x / (x²+1) then g'(x) = 0 ---> x = 0

Second, find g(x ) for all x such as:

x = -1 ---> g(-1) = 12 - Ln( 2)

x = 0 ----> g(0) = 12 -Ln(1) = 12 - 0 = 12

x = 5 ----> g(5) = 12 - Ln (26)

Third, which of g(x) is biggest see !! 12 so max value of g(x) = 12

I hope it is useful,please let me know if you have any question.

Minoo

Answer 2

Since our function is g(x)=12-ln(x²+1) and we wish to find the value of x on the given interval which maximizes g(x) we need to find the value of x which will make the value of ln(x²+1) as small as possible (since subtracting a smaller number is equivalent to addition). Now notice that ln(u) is an increasing function of u, which means that as u gets bigger, so does ln(u). This means that in order to make ln(x²+1) as small as possible we need to make x²+1 as small as possible, and thus to make x² as small as possible on the given interval. Clearly x² is minimized when x=0, so this is the value of x for which g(x) is maximized on the given interval.

Answer 3

0=g'(x) = (-2x)/ (x^2+1)

so it happens at x=0

(0,12)