Let f: R→R be given, such that ∀x: f(x)·f'(x)≥0. Show that if f(0)= 0, then ∀x∈(-∞,0): f(x)= 0.

This statement can be proven using a definite integral.

Expert Answers

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Consider that `( ( f ( x ) ) ^ 2 ) ' = 2 f ( x ) f ' ( x ) , ` which is given to be non-negative. Let's integrate this function from some negative `x ` to zero:

`int_( x )^( 0 ) ( 2 f ( x ) f ' ( x ) ) dx = int_( x )^( 0 ) ( ( ( f ( x ) ) ^ 2 ) ' ) dx = f ( ( 0 ) ) ^ 2 - ( f ( x ) ) ^ 2 = - ( f ( x ) ) ^ 2 .`

But the expression from which we started, `int_( x )^( 0 ) ( 2 f ( x ) f ' ( x ) ) dx , ` is non-negative because the function under the integral is non-negative. On the other hand, the last expression, `- ( f ( x ) ) ^ 2 , ` is obviously non-positive.

This situation has only one resolution—that both quantities are zero. It is what we need to prove, because `- ( f ( x ) ) ^ 2 = 0 ` for any `x lt 0 ` implies `f ( x ) = 0 ` for any `x lt 0 .`

If you need proof that the integral of a non-negative function is non-negative, consider integral sums.

Last Updated by eNotes Editorial on February 16, 2021
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