Question

Which of the following is not true for $f(x) = \min(x, x^2)$?

• A. continuous everywhere
• B. $Rf'(0) = 1$
• C. not differentiable at $x = 0$ and $x = 1$
• D. $Rf'(1) = 1$

Answer 1

Answer: (B)

$Rf'(0) = 1$

Answer 2

The function is defined as $f(x) = \min(x, x^2)$. We first determine the piecewise form of $f(x)$. The equality $x = x^2$ holds when $x^2 - x = 0$, which factors as $x(x-1) = 0$. Thus, the intersection points are $x=0$ and $x=1$.

We analyze the inequality $x \le x^2$, which is equivalent to $x^2 - x \ge 0$, or $x(x-1) \ge 0$. This inequality holds for $x \in (-\infty, 0] \cup [1, \infty)$. Therefore, $f(x) = x$ for $x \le 0$ or $x \ge 1$.

The inequality $x > x^2$ holds when $x^2 - x < 0$, or $x(x-1) < 0$. This inequality holds for $x \in (0, 1)$. Therefore, $f(x) = x^2$ for $0 < x < 1$.

The piecewise definition of $f(x)$ is:

$$f(x) = \begin{cases} x & \text{if } x \le 0 \\ x^2 & \text{if } 0 < x < 1 \\ x & \text{if } x \ge 1 \end{cases}$$

Now, let's evaluate each statement:

1. Continuous everywhere: We check continuity at the transition points $x=0$ and $x=1$. At $x=0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0$. $f(0) = 0$. Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$, $f(x)$ is continuous at $x=0$.

   At $x=1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x = 1$. $f(1) = 1$. Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$, $f(x)$ is continuous at $x=1$. Since $f(x)$ is defined by polynomials ($x$ or $x^2$) in the intervals $(-\infty, 0)$, $(0, 1)$, and $(1, \infty)$, it is continuous there. Thus, $f(x)$ is continuous everywhere. This statement is true.

2. $Rf'(0) = 1$: $Rf'(0)$ is the right-hand derivative of $f(x)$ at $x=0$. $Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$. Since $f(0) = 0$ and for small $h > 0$, $f(h) = h^2$ (as $h \in (0, 1)$), we have: $Rf'(0) = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0.$ The statement claims $Rf'(0) = 1$, which is false. This statement is not true.

3. Not differentiable at $x = 0$ and $x = 1$: At $x=0$: The left-hand derivative is $Lf'(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}$. For $h < 0$, $f(h) = h$. $Lf'(0) = \lim_{h \to 0^-} \frac{h - 0}{h} = \lim_{h \to 0^-} 1 = 1.$ Since $Lf'(0) = 1$ and $Rf'(0) = 0$, $f(x)$ is not differentiable at $x=0$.

   At $x=1$: The left-hand derivative is $Lf'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$. For small $h < 0$, $1+h \in (0, 1)$, so $f(1+h) = (1+h)^2$. $f(1)=1$. $Lf'(1) = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{1 + 2h + h^2 - 1}{h} = \lim_{h \to 0^-} (2+h) = 2.$ The right-hand derivative is $Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$. For small $h > 0$, $1+h \ge 1$, so $f(1+h) = 1+h$. $Rf'(1) = \lim_{h \to 0^+} \frac{(1+h) - 1}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1.$ Since $Lf'(1) = 2$ and $Rf'(1) = 1$, $f(x)$ is not differentiable at $x=1$. Therefore, $f(x)$ is not differentiable at $x=0$ and $x=1$. This statement is true.

4. $Rf'(1) = 1$: From our calculation in point 3, we found $Rf'(1) = 1$. This statement is true.

The question asks which statement is not true. Only statement 2, "$Rf'(0) = 1$", is false. The correct value for the right-hand derivative at $x=0$ is $0$.