# Determine whether the theorem is properly applied to the function below. If not, indicate why not. (Pick A-E) f(x)=xln(|x|+1). Rolle's theorem applied on the interval [-1,1] implies that there is a point in (-4,4) where f '(x)=0. A. There are point(s) on the interval [a,b] where f is not continous. B. There are point(s) on the interval (a,b) where f is not differentiable. C. f(a) does not equal f(b) for any distinct a,b in the interval [-1,1]. D. The theorem is properly applied. E. None of these Let's prove that f is increasing on [-1,1] hence there are no a and b (a!=b) such that f(a)=f(b)

On [-1,0], x<0 and ln(|x|+1)>0

For any a<b<0

|a|>|b|>0

ln(|a|+1)>ln(|b|+1)>0

-a>-b>0

Multiply the inequalities with positive terms.

-aln(|a|+1)>-bln(|b|+1)>0

Therefore

aln(|a|+1)<bln(|b|+1)<0

and f is increasing on [-1,0]

On (0,1]

x is a positive increasing function.

|x| is an increasing function therefore 1+|x| is an increasing function and 1+|x|>1

Hence ln(1+|x|) is a positive increasing function. The product of 2 positive increasing functions is an increasing function.

f is an increasing function on (0,1]

In conclusion, f is increasing on [-1,1] therefore there are no `a!=b` such that f(a)=f(b).

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