The number of integral elements in th range of

f(x)= ( x ( x² - 1 ) ) / (x⁴ - x² + 1) will be …........?

To determine the number of integral elements in the range of the function \( f(x) = \frac{x(x^2 - 1)}{x^4 - x^2 + 1} \), we need to analyze the behavior of this function. This involves finding its critical points, understanding its limits, and identifying any asymptotic behavior. Let's break it down step by step.

Understanding the Function

The function can be rewritten as:

f(x) = \frac{x^3 - x}{x^4 - x^2 + 1}

Here, the numerator \( x^3 - x \) is a cubic polynomial, and the denominator \( x^4 - x^2 + 1 \) is a quartic polynomial. The degree of the numerator is less than the degree of the denominator, which suggests that the function may have horizontal asymptotes as \( x \) approaches infinity or negative infinity.

Finding Limits

Let's evaluate the limits of \( f(x) \) as \( x \) approaches positive and negative infinity:

• As \( x \to \infty \):

  Both the numerator and denominator are dominated by their highest degree terms. Thus, we have:

  f(x) \approx \frac{x^3}{x^4} = \frac{1}{x} \to 0

• As \( x \to -\infty \):

  Similarly, we find:

  f(x) \approx \frac{x^3}{x^4} = \frac{1}{x} \to 0

Critical Points and Behavior

Next, we need to find the critical points by taking the derivative of \( f(x) \) and setting it to zero. This will help us identify local maxima and minima:

Using the quotient rule:

f'(x) = \frac{(x^4 - x^2 + 1)(3x^2 - 1) - (x^3 - x)(4x^3 - 2x)}{(x^4 - x^2 + 1)^2}

Setting the numerator equal to zero will give us the critical points. However, solving this can be complex, so we can also analyze the function's behavior at specific points.

Evaluating Specific Values

Let's evaluate \( f(x) \) at some key points:

• At \( x = 0 \):

  f(0) = \frac{0(0^2 - 1)}{0^4 - 0^2 + 1} = 0

• At \( x = 1 \):

  f(1) = \frac{1(1^2 - 1)}{1^4 - 1^2 + 1} = 0

• At \( x = -1 \):

  f(-1) = \frac{-1((-1)^2 - 1)}{(-1)^4 - (-1)^2 + 1} = 0

• At \( x = 2 \):

  f(2) = \frac{2(2^2 - 1)}{2^4 - 2^2 + 1} = \frac{2(3)}{13} = \frac{6}{13}

• At \( x = -2 \):

  f(-2) = \frac{-2(3)}{13} = -\frac{6}{13}

Range Analysis

From our evaluations, we see that \( f(x) \) takes values around 0 and approaches it as \( x \) goes to infinity or negative infinity. The function is continuous, and we can expect it to take on values between its local maxima and minima. Since we have established that the function approaches 0 and takes values like \( \frac{6}{13} \) and \( -\frac{6}{13} \), we can conclude that the function's range is likely between these values.

Integral Elements in the Range

Given that the function approaches 0 and takes values close to \( \pm \frac{6}{13} \), we can identify the integral elements in the range:

• Since \( f(x) \) can take values between \( -\frac{6}{13} \) and \( \frac{6}{13} \), the only integer that lies within this interval is 0.

Thus, the number of integral elements in the range of \( f(x) \) is 1.