Question

In the interval (-2\(\pi\),0), the function f(x)=sin(\(\frac{1}{x^3}\)).

• A. never changes sign
• B. changes sign only once
• C. changes sign more than once but finitely many times.
• D. changes sign infinitely many times

Answer 1

The Correct Option is D

We are given the function: $f(x) = \sin\left(\frac{1}{x^3}\right)$ and asked about its sign-changing behavior in the interval $(-2\pi, 0)$.

Let us analyze the behavior of $\frac{1}{x^3}$ as $x$ approaches 0 from the left:

• As $x \to 0^-$, we have $x^3 \to 0^-$.
• So, $\frac{1}{x^3} \to -\infty$. 

Now consider the composition $f(x) = \sin\left(\frac{1}{x^3}\right)$. Since $\frac{1}{x^3}$ becomes large in magnitude and negative near $x = 0$, the input to the sine function becomes a large negative number.

We know that the sine function, $\sin(\theta)$, is periodic and oscillates between $-1$ and $1$. It crosses zero at multiples of $\pi$:

$\sin(\theta) = 0$ when $\theta = n\pi$ for any integer $n$.

So, as $x \to 0^-$, $\frac{1}{x^3}$ passes through infinitely many multiples of $\pi$ in the negative direction. That means:

• $\sin\left(\frac{1}{x^3}\right)$ will be $0$ for infinitely many values of $x$.
• Between any two zeros of the sine function, it takes on both positive and negative values.

Therefore, $f(x)$ changes sign infinitely many times as $x \to 0^-$.

Since $x$ never equals $0$ (we're only looking in the interval $(-2\pi, 0)$), this oscillation continues endlessly as we approach $0$ from the left.

Final conclusion: The function $f(x) = \sin\left(\frac{1}{x^3}\right)$ changes sign infinitely many times in the interval $(-2\pi, 0)$.

Correct option: (D) changes sign infinitely many times

Answer 2

Sign Changes of $f(x) = \sin\left(\frac{1}{x^3}\right)$ in $(-2\pi, 0)$ 

Step 1: The sine function $\sin(u)$ changes sign whenever $u = k\pi$, where $k$ is any integer.

Step 2: For $f(x) = \sin\left(\frac{1}{x^3}\right)$ to change sign, we want:

$\frac{1}{x^3} = k\pi \Rightarrow x^3 = \frac{1}{k\pi} \Rightarrow x = \frac{1}{\sqrt[3]{k\pi}}$

Step 3: We're looking in the interval $(-2\pi, 0)$, so $x$ must be negative. That means $k$ must also be negative. Let $k = -n$, where $n \in \mathbb{N}$:

$x = \frac{1}{\sqrt[3]{-n\pi}} = -\frac{1}{\sqrt[3]{n\pi}}$

Step 4: We want values of $x$ such that $-2\pi < x < 0$, which gives:

$-2\pi < -\frac{1}{\sqrt[3]{n\pi}} < 0$

Multiplying by $-1$ (and flipping inequality):

$0 < \frac{1}{\sqrt[3]{n\pi}} < 2\pi$

Taking reciprocal and flipping:

$\frac{1}{2\pi} < \sqrt[3]{n\pi}$

Cubing both sides:

$\frac{1}{8\pi^3} < n\pi \Rightarrow \frac{1}{8\pi^4} < n$

Step 5: Since there are infinitely many positive integers $n$ satisfying this, there are infinitely many corresponding $x$-values in $(-2\pi, 0)$ where $\sin\left(\frac{1}{x^3}\right)$ changes sign.

Final Answer: The function changes sign infinitely many times in the interval $(-2\pi, 0)$.