1. **State the problem:** We want to find the number of points where the function $$f(x) = \begin{cases} \min(1, x^2, x^3), & x < 1 \\ \min(x^3, 3x - 2), & x \geq 1 \end{cases}$$ is not differentiable. 2. **Analyze the function for $x < 1$:** We have $f(x) = \min(1, x^2, x^3)$. - For $x < 1$, compare $1$, $x^2$, and $x^3$. - Since $x^2 \geq 0$ and $x^3$ can be negative for $x<0$, the minimum depends on $x$. 3. **Find where the minimum changes for $x < 1$:** - Compare $1$ and $x^2$: $x^2 \leq 1$ for all $x$. - Compare $x^2$ and $x^3$: Solve $x^2 = x^3 \Rightarrow x^2(1 - x) = 0 \Rightarrow x=0$ or $x=1$. - For $x<0$, $x^3 < x^2$, so minimum is $x^3$. - For $0 \leq x < 1$, $x^2 < x^3$, so minimum is $x^2$. - Also, compare $1$ and $x^3$: Since $x^3 < 1$ for $x<1$, $1$ is not minimum except possibly at $x=1$. Thus, for $x<1$: - $x < 0$: $f(x) = x^3$ - $0 \leq x < 1$: $f(x) = x^2$ 4. **Analyze the function for $x \geq 1$:** We have $f(x) = \min(x^3, 3x - 2)$. - Find where $x^3 = 3x - 2$: $$x^3 - 3x + 2 = 0$$ - Factor: $$(x - 1)^2 (x + 2) = 0$$ - Roots at $x=1$ (double root) and $x=-2$ (not in domain $x \geq 1$). For $x > 1$, check which is smaller: - At $x=2$, $x^3=8$, $3x-2=4$, so minimum is $3x-2$. - At $x=1$, both equal 1. So for $x \geq 1$, - $f(x) = x^3$ at $x=1$ - $f(x) = 3x - 2$ for $x > 1$ 5. **Check differentiability at critical points:** - At $x=0$, function changes from $x^3$ to $x^2$. Left derivative: $f'(0^-) = 3(0)^2 = 0$ Right derivative: $f'(0^+) = 2(0) = 0$ Derivatives match, so differentiable at $x=0$. - At $x=1$, function changes from $x^2$ (for $x<1$) to $x^3$ (at $x=1$) and then to $3x-2$ (for $x>1$). Left derivative: $f'(1^-) = 2(1) = 2$ Right derivative: $f'(1^+) = \frac{d}{dx}(3x-2)\big|_{x=1} = 3$ Derivatives do not match, so not differentiable at $x=1$. - Check if $f$ is differentiable at points where minimum changes inside $x<1$: At $x=0$, already checked. No other change points inside $x<1$. 6. **Check differentiability at points where minimum switches for $x \geq 1$:** - At $x=1$, already checked. - No other points since $x=1$ is the only root. 7. **Conclusion:** The function is not differentiable only at $x=1$. **Final answer:** There is 1 point where $f$ is not differentiable. **Answer choice:** (2) 1