91Ó°ÊÓ

Consider the function \(g(x)=|x-1|\). The kink happens when the inside of the absolute value (or the argument) is equal to zero, i.e., \(x-1=0\). Solving for x, we get \(x=1\). This is the point where the absolute value function takes a "sharp turn".

We have proved that the function \(g(x)=|x-1|\) has a kink and is non-differentiable at \(x=1\). Now, to check the differentiability of the entire function \(f(x)=(x+1)^{23}+|x-1|^{\sqrt{3}}\), at \(x=1\), we need to find the left and right-hand derivatives. The left-hand derivative at \(x=1\) is given by: \(\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{((1+h)+1)^{23} + |(1+h)-1|^{\sqrt{3}} - 2^{23}}{h}\) For \(h<0\), we have that \(1+h<1\), and thus the absolute value in the limit becomes \((1 - (1+h))^{\sqrt{3}} = (-h)^{\sqrt{3}} = h^{\sqrt{3}}\). Substitute this back into the limit: \(\lim_{h \to 0^-} \frac{((1+h)+1)^{23} + h^{\sqrt{3}} - 2^{23}}{h}\) Since \((1+h)+1 = 2+h\), and \(2^{23}\) is a constant in terms of h, we can rewrite this as: \(\lim_{h \to 0^-} \frac{(2+h)^{23} - 2^{23} + h^{\sqrt{3}}}{h}\) Similarly, the right-hand derivative at \(x=1\) is given by: \(\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{((1+h)+1)^{23} + |(1+h)-1|^{\sqrt{3}} - 2^{23}}{h}\) For \(h>0\), the absolute value in the limit becomes \((1+h-1)^{\sqrt{3}} = h^{\sqrt{3}}\). Substitute this back into the limit: \(\lim_{h \to 0^+} \frac{(2+h)^{23} - 2^{23} + h^{\sqrt{3}}}{h}\) We can see that the left-hand derivative and the right-hand derivative are of the same form.

Using L'Hôpital's rule, we can find each limit one by one: \(\lim_{h \to 0^-} \frac{(2+h)^{23} - 2^{23} + h^{\sqrt{3}}}{h} = \lim_{h \to 0^-} \frac{23(2+h)^{22} + \sqrt{3}h^{\sqrt{3} - 1}}{1}\) Similarly, for the right-hand derivative: \(\lim_{h \to 0^+} \frac{(2+h)^{23} - 2^{23} + h^{\sqrt{3}}}{h} = \lim_{h \to 0^+} \frac{23(2+h)^{22} + \sqrt{3}h^{\sqrt{3} - 1}}{1}\) As we can see, both the left-hand derivative and the right-hand derivative are equal when h approaches to 0. Thus, the function \(f(x)\) is differentiable at x = 1.

Since both the left and right-hand derivatives exist and are equal at the kink point \(x=1\), the function \(f(x)\) is differentiable at this point. This means there are no non-differentiable points for the function \(f(x)\). Hence, the correct answer is: (D) none