Question

Prove that: (image attached)

Answer 1

x belongs to [-1;0]

Answer 2

and \[f(x) = x ^{2} \times e ^{x}\]

Answer 3

g(x) = f(x) + f(x^2) is inverse in [-1; 0]
We can differentiate g(x) and consider the sign of g'(x).

Answer 4

yeah...as jean said u just need find critical points on the given interval...and since\[g(x)=f(x)+f(x^2)\]is continous check for the values of the g(x) in the interval bounds

Answer 5

so i should use the first derivative of the g(x)... jean seand that the function is inverse on the given interval. how's that?

Answer 6

said*

Answer 7

i dont know what he/she means!

Answer 8

\[g(x)=x ^{2}timese ^{x}+x ^{4}timese ^{x ^{2}}\]

Answer 9

sr, i mean, just find g'(x) and check if g'(x) < 0

Answer 10

mm. and if g(x) - descending function => it is smaller than \[\frac{ e ^{2}+1 }{ e }\]

Answer 11

yeah, g(0) <= g(x) <= g(-1)

Answer 12

right. at mathematical analysis there are lots of inequalities of this type. so the main method of solving them is this above mentioned....

Answer 13

thanks.

Answer 14

welcome...