Let Q be a constant so that fx Q 1x3 for the second derivat

Let Q be a constant so that f\'\'(x) = Q (1+x)3 for the second derivative f\'\'(x) of f(x) = x2 1+x. Then the third decimal of ln(3+lQl) is

Solution

i think you mistyped f(x) = x2 1+x because if that is x^(21) then you wont get (1+x)^3 in f\'\'(x) as given

but that can be obtained using f(x)=(1+x)^5

f\'(x)=5(1+x)^4

f\'\'(x)=20(1+x)^3

now comparing with f\'\'(x) = Q (1+x)^3 gives

Q=20

now plug it into ln(3+lQl)

=ln(3+l20l)=ln(3+20)=ln(23)=3.13549421593

so third decimal of ln(3+lQl) is 5