MHB What are the implications of x=0 in terms of function derivability?

[Noismaker]

View attachment 6622

Thank you

 

[HallsofIvy]

I think "differentiability" would be a better word than "derivability" since that might be confused with the ability to "derive" a formula. In any case, the first thing I would do is remove the "absolute value" by looking at x positive or negative separately.

If x is positive, then f(x)= x^x and, taking the logarithm of both sides, log(f(x))= x log(x). Differentiating, f'(x)/f(x)= log(x)+ 1 so that f'(x)= (log(x)+ 1)f(x)= (log(x)+ 1)x^x. That exists for all positive x so f(x) is differentiable for all positive x.

If x is negative, make the substitution y= -x. |x|= -x= y so that x^x= y^{-y}. Again take the logarithm of both sides of f(y)= y^{-y}, log(f(y))= -ylog(y). Differentiating, f&#039;(y)/f(y)= -log(y)- 1 so that f&#039;(y)= -f(y)(log(y)+ 1)= -y^y(log(y)+ 1) and, since dy/dx= 0, f&amp;#039;(x)= (-x)^x(log(|x|)+ 1). That exists for all negative x so f(x) is differentiable for all negative x.<br /> <br /> Finally, look at x= 0. The derivative of f(x) at x= 0 is \lim_{h\to 0} \frac{f(h)- f(0)}{h}= \lim_{h\to 0}\frac{h^h- 1}{h}. Does that limit exist?

 

[Noismaker]

1) View attachment 6623
2) View attachment 6624
3) View attachment 6625

so in x=0 we have a point of non-differentiation, cusp, angular point, or flattened to vertical tangent.