1) find x' then find x''(t)=-10 . x'=- and x(1)=6
how do you start?
find absolute max and min of f(x) = x/lnx from [sqrt e, e^4) I got ln(x)-1/ln^2(x) =0
x= e, 0 , 1
if x''(t)=-10
then x' = -10t + c
you then have a typo in "x'=- " , just for arguments sake, suppose it was
x'(2) = - 3
the -3 = -10(2) + c
c = 17
x'(t) = -10t + 17
x(t) = -5t^2 + 17t + k
given: x(1)=6
6 = -5(1) + 17(1) + k
k = -6
x(t) = -5T^2 + 17t - 6
make the necessary changes once you have determined what
x'=- means
for your 2nd:
f(x) = x/lnx
f'(x) = (lnx - x/x)/(lnx)^2 = 0 for max/min
lnx - 1 = 0
lnx = 1
x = e
f(e) = e/lne = e = appr 2.718..
for endpoints:
f(√e) = 3.297..
f(e^4) = 13.649...
max is f(e^4)
min is e