# Explain why is 1-x^2 less than 1/e^x^2?

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### 1 Answer

Write the inequality `1-x^2 lt 1/(e^(x^2))`

Subtracting `1/(e^(x^2))` both sides yields:

`1-x^2 - 1/(e^(x^2)) lt 0`

`` Use the function `1-x^2 - 1/(e^(x^2)) = f(x)`

`` You need to prove that f(x)<0.

Use derivatives to prove that the function decreases on R.

Find the derivative of f(x) using the quotient and chain rules at the second term.

`f'(x) = -2x - (-2x*e^(x^2))/(e^(x^2))^2`

Reducing `e^(x^2) ` yields:

`f'(x) = -2x + 2x/(e^(x^2))`

Notice the common factor 2x:

`f'(x) = 2x (-1 + 1/(e^(x^2)))`

Notice that `1 gt 1/(e^(x^2)) =gt -1 + 1/(e^(x^2)) lt 0`

If x<0 => f'(x) > 0

If x>0 => f'(x) < 0

**The function decreases on the interval `(0 ; +oo)` =>`1-x^2 lt 1/(e^(x^2))` if x `in` R.**