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
1 Answer | Add Yours
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
Multiply the inequalities with positive terms.
and f is increasing on [-1,0]
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).
We’ve answered 318,983 questions. We can answer yours, too.Ask a question