Question

Let $f(x)=\begin{cases} (x-1)(x-n) & x \leq n \\ x^{2}-n & x > n \end{cases}$, then identify correct statement(s) ($n \in I$)

• A. $\forall$ n ≤ 0 f(x) has local minima at x = n
• B. $\forall$ n ≥ 0 f(x) has local minima at x = n
• C. $\forall$ n > 0 f(x) is strictly increasing at x = n
• D. f(x) has always a local minima $\forall$ n$\in$I

Answer 1

Answer: (A)

A

Answer 2

The function is given by: $f(x)=\begin{cases} (x-1)(x-n) & x \leq n \\ x^{2}-n & x > n \end{cases}$ where $n \in I$ (n is an integer).

First, let's evaluate $f(x)$ at $x=n$: $f(n) = (n-1)(n-n) = (n-1) \cdot 0 = 0$.

For $f(x)$ to have a local minimum at $x=n$, there must exist an open interval $(n-\delta, n+\delta)$ such that $f(x) \geq f(n)$ for all $x \in (n-\delta, n+\delta)$. Since $f(n)=0$, we need $f(x) \geq 0$ in this interval.

Let's analyze the two parts of the function around $x=n$:

1. For $x > n$ (and close to $n$): $f(x) = x^2 - n$. We need $x^2 - n \geq 0$. Let $x = n+\epsilon$ for a small $\epsilon > 0$. $f(n+\epsilon) = (n+\epsilon)^2 - n = n^2 + 2n\epsilon + \epsilon^2 - n = (n^2-n) + (2n\epsilon + \epsilon^2)$. We need $(n^2-n) + (2n\epsilon + \epsilon^2) \geq 0$ for small $\epsilon > 0$. If $n^2-n > 0$, this condition is satisfied for small $\epsilon$. ($n(n-1)>0 \implies n \in (-\infty, 0) \cup (1, \infty)$). If $n^2-n = 0$, i.e., $n=0$ or $n=1$, then $f(n+\epsilon) = 2n\epsilon + \epsilon^2$. For $n=0$, $f(\epsilon) = \epsilon^2 \geq 0$. For $n=1$, $f(1+\epsilon) = 2\epsilon + \epsilon^2 \geq 0$ for small $\epsilon > 0$. So, for $x > n$ and close to $n$, $f(x) \geq f(n)$ is always true for any integer $n$.

2. For $x \leq n$ (and close to $n$): $f(x) = (x-1)(x-n)$. We need $(x-1)(x-n) \geq 0$. Let $x = n-\epsilon$ for a small $\epsilon > 0$. $f(n-\epsilon) = ((n-\epsilon)-1)((n-\epsilon)-n) = (n-1-\epsilon)(-\epsilon) = -\epsilon(n-1-\epsilon)$. For $f(n-\epsilon) \geq 0$, we need $-\epsilon(n-1-\epsilon) \geq 0$. Since $\epsilon > 0$, we must have $-(n-1-\epsilon) \geq 0$, which implies $n-1-\epsilon \leq 0$. So, $n-1 \leq \epsilon$.

   This condition must hold for some sufficiently small $\epsilon > 0$.

   • If $n-1 \leq 0$ (i.e., $n \leq 1$), then for any small $\epsilon > 0$, we can choose $\epsilon$ such that $n-1 \leq \epsilon$. In this case, $n-1-\epsilon \leq 0$, so $-\epsilon(n-1-\epsilon) \geq 0$. Thus, if $n \leq 1$, $f(x) \geq f(n)$ for $x < n$ and close to $n$.
   • If $n-1 > 0$ (i.e., $n > 1$), then we can choose $\epsilon$ such that $0 < \epsilon < n-1$. For such an $\epsilon$, $n-1-\epsilon > 0$. In this case, $f(n-\epsilon) = -\epsilon(n-1-\epsilon) < 0$. This means $f(x) < f(n)$ for $x$ slightly less than $n$. Therefore, if $n > 1$, $f(x)$ does not have a local minimum at $x=n$.

Combining the results from both sides, $f(x)$ has a local minimum at $x=n$ if and only if $n \leq 1$. Since $n \in I$, this includes $n = 1, 0, -1, -2, \ldots$.

Now let's evaluate the given options:

(A) $\forall$ n ≤ 0 f(x) has local minima at x = n. This statement is true. If $n \leq 0$, then $n$ is certainly $\leq 1$. So, $f(x)$ has a local minimum at $x=n$.

(B) $\forall$ n ≥ 0 f(x) has local minima at x = n. This statement is false. For example, if $n=2$, then $n \geq 0$ but $n > 1$. Our analysis shows that for $n > 1$, $f(x)$ does not have a local minimum at $x=n$.

(C) $\forall$ n > 0 f(x) is strictly increasing at x = n. For $f(x)$ to be strictly increasing at $x=n$, we would need $f(x_1) < f(x_2)$ for $x_1 < x_2$ in a neighborhood of $n$. Consider $n=1$. We found $f(x)$ has a local minimum at $x=1$. A function with a local minimum is not strictly increasing at that point. Consider $n > 1$. For $x \in (1, n)$, $f(x) = (x-1)(x-n)$. Since $x-1 > 0$ and $x-n < 0$, $f(x) < 0$. Since $f(n)=0$, we have $f(x) < f(n)$ for $x$ approaching $n$ from the left. This means $f(x)$ is decreasing as $x \to n^-$. Therefore, $f(x)$ is not strictly increasing at $x=n$ for $n > 0$. This statement is false.

(D) f(x) has always a local minima $\forall$ n$\in$I. This statement is false. As shown, $f(x)$ only has a local minimum at $x=n$ if $n \leq 1$. For $n > 1$, it does not.

The only correct statement is (A).