Extrema shift 2

Calculus Level pending

For a function f ( x ) = 1 − x 1 + x 2 e x f(x)=\frac{1-x}{1+x^{2}}e^{x} , there exist x 1 , x 2 ( x 1 ≠ x 2 ) x_{1},x_{2} (x_{1}≠x_{2}) such that f ( x 1 ) = f ( x 2 ) f(x_{1})=f(x_{2}) .

What is always true for x 1 + x 2 x_{1} + x_{2} ?

x 1 + x 2 > 0 x_{1} + x_{2} > 0 x 1 + x 2 < 0 x_{1} + x_{2} < 0 x 1 + x 2 = 0 x_{1} + x_{2} = 0 Cannot be determined

Alice Smith by Alice Smith , 20, China

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We see that f(x) is a negatively skewed function, with f(-infinity) =f(1)=0, f(0)=1 is the maximum value of the function. Therefore by Rolle's theorem, x1 is negative and x2 is positive, and due to the nature of the skewness, x1+x2<0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...