${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$ where x ϵ RWhat is the minimum value of f(x)?

Question

${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$ where x &varepsilon; RWhat is the minimum value of f(x)?

Prepp Team · Accepted Answer

Finding the Minimum Value of the Function f(x)

The problem asks us to find the minimum value of the function given by:

${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$, where ${\rm{x}} \in {\rm{R}}$.

We need to find the lowest possible value that this function can take for any real number x.

Analyzing the Function Algebraically

Let's rewrite the function f(x) in a different form to make its behavior clearer. We can perform polynomial division or algebraic manipulation:

${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}} = \frac{{{{\rm{x}}^2} + 1 - 2}}{{{{\rm{x}}^2} + 1}}$

Now, we can split this into two terms:

${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} + 1}}{{{{\rm{x}}^2} + 1}} - \frac{2}{{{{\rm{x}}^2} + 1}}$

${\rm{f}}\left( {\rm{x}} \right) = 1 - \frac{2}{{{{\rm{x}}^2} + 1}}$

Now let's analyze the term $\frac{2}{{{{\rm{x}}^2} + 1}}$:

The term ${\rm{x}}^2$ is always non-negative for any real number x (${\rm{x}}^2 \ge 0$).
Therefore, ${\rm{x}}^2 + 1$ is always greater than or equal to 1 (${\rm{x}}^2 + 1 \ge 1$).

Consider the fraction $\frac{2}{{{{\rm{x}}^2} + 1}}$. Since the numerator is a positive constant (2) and the denominator ${\rm{x}}^2 + 1$ is always positive and greater than or equal to 1, the fraction will always be positive. The value of this fraction depends on the value of ${\rm{x}}^2 + 1$.

To maximize the fraction $\frac{2}{{{{\rm{x}}^2} + 1}}$, we need to minimize the denominator ${\rm{x}}^2 + 1$. The minimum value of ${\rm{x}}^2 + 1$ is 1, which occurs when ${\rm{x}}^2 = 0$, i.e., when ${\rm{x}} = 0$.
The maximum value of $\frac{2}{{{{\rm{x}}^2} + 1}}$ is $\frac{2}{1} = 2$, which occurs at ${\rm{x}} = 0$.
As $|{\rm{x}}|$ increases (approaches infinity), ${\rm{x}}^2 + 1$ increases and approaches infinity. Thus, $\frac{2}{{{{\rm{x}}^2} + 1}}$ decreases and approaches 0.

Now let's look at the expression for ${\rm{f}}({\rm{x}})$ again: ${\rm{f}}\left( {\rm{x}} \right) = 1 - \frac{2}{{{{\rm{x}}^2} + 1}}$.

To find the minimum value of ${\rm{f}}({\rm{x}})$, we need to subtract the largest possible value from 1. The maximum value of $\frac{2}{{{{\rm{x}}^2} + 1}}$ is 2, and this occurs at ${\rm{x}} = 0$.

So, the minimum value of ${\rm{f}}({\rm{x}})$ occurs at ${\rm{x}} = 0$.

Let's calculate the value of ${\rm{f}}({\rm{x}})$ at ${\rm{x}} = 0$:

${\rm{f}}\left( 0 \right) = 1 - \frac{2}{{{0^2} + 1}} = 1 - \frac{2}{1} = 1 - 2 = -1$

Alternatively, using the original expression:

${\rm{f}}\left( 0 \right) = \frac{{{0^2} - 1}}{{{0^2} + 1}} = \frac{{ - 1}}{1} = -1$

For any other value of x, ${\rm{x}}^2 + 1 > 1$, so $0 < \frac{2}{{{{\rm{x}}^2} + 1}} < 2$. This means ${\rm{f}}({\rm{x}}) = 1 - \frac{2}{{{{\rm{x}}^2} + 1}}$ will be greater than $1 - 2 = -1$ but less than $1 - 0 = 1$. The function approaches 1 as $|{\rm{x}}|$ gets very large, but it never reaches 1.

Thus, the minimum value of the function ${\rm{f}}({\rm{x}})$ is -1.

Using Calculus (for completeness)

We can also find the minimum value using calculus by finding the derivative and critical points.

${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$

Using the quotient rule, ${\rm{f}}'({\rm{x}}) = \frac{{{\rm{d}}}}{{{\rm{d}}x}}\left( {\frac{{{\rm{u}}}}{{\rm{v}}}} \right) = \frac{{{\rm{u}}'{\rm{v}} - {\rm{uv}}'}}{{{{\rm{v}}^2}}}$, where ${\rm{u}} = {\rm{x}}^2 - 1$ and ${\rm{v}} = {\rm{x}}^2 + 1$.

${\rm{u}}' = 2{\rm{x}}$ and ${\rm{v}}' = 2{\rm{x}}$.

${\rm{f}}'({\rm{x}}) = \frac{{(2{\rm{x}})({\rm{x}}^2 + 1) - ({\rm{x}}^2 - 1)(2{\rm{x}})}}{{({\rm{x}}^2 + 1)^2}}$

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

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

${\rm{f}}'({\rm{x}}) = \frac{{4{\rm{x}}}}{{({\rm{x}}^2 + 1)^2}}$

To find critical points, set ${\rm{f}}'({\rm{x}}) = 0$:

$\frac{{4{\rm{x}}}}{{({\rm{x}}^2 + 1)^2}} = 0$

Since $({\rm{x}}^2 + 1)^2$ is never zero for real x, we only need to consider the numerator:

$4{\rm{x}} = 0 \Rightarrow {\rm{x}} = 0$

The only critical point is ${\rm{x}} = 0$. To determine if this is a minimum or maximum, we can check the sign of ${\rm{f}}'({\rm{x}})$ around ${\rm{x}}=0$.

For ${\rm{x}} < 0$, $4{\rm{x}}$ is negative, $({\rm{x}}^2 + 1)^2$ is positive. So, ${\rm{f}}'({\rm{x}}) < 0$. The function is decreasing.
For ${\rm{x}} > 0$, $4{\rm{x}}$ is positive, $({\rm{x}}^2 + 1)^2$ is positive. So, ${\rm{f}}'({\rm{x}}) > 0$. The function is increasing.

Since the function changes from decreasing to increasing at ${\rm{x}} = 0$, there is a local minimum at ${\rm{x}} = 0$. Evaluating the function at ${\rm{x}} = 0$:

${\rm{f}}\left( 0 \right) = \frac{{{0^2} - 1}}{{{0^2} + 1}} = \frac{{ - 1}}{1} = -1$

Considering the behavior as ${\rm{x}} \to \pm \infty$, ${\rm{f}}({\rm{x}}) \to 1$. The lowest point the function reaches is -1.

Conclusion

Both the algebraic analysis and the calculus method confirm that the minimum value of the function ${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$ is -1.

Method	Process	Result
Algebraic Analysis	Rewrite ${\rm{f}}({\rm{x}}) = 1 - \frac{2}{{{{\rm{x}}^2} + 1}}$. Analyze minimum of denominator ${\rm{x}}^2+1$ to find maximum of $\frac{2}{{{{\rm{x}}^2} + 1}}$, which gives minimum of $1 - \frac{2}{{{{\rm{x}}^2} + 1}}$.	Minimum value is -1 at x=0.
Calculus	Find the derivative ${\rm{f}}'({\rm{x}})$, set ${\rm{f}}'({\rm{x}}) = 0$ to find critical points. Use the first derivative test to classify the critical point.	Critical point at x=0, which is a local minimum. ${\rm{f}}(0) = -1$.

Revision Table: Minimum Value of a Function

Concept	Explanation	Application to f(x)
Minimum Value	The smallest value in the range of the function.	Seeking the lowest possible output of ${\rm{f}}({\rm{x}})$.
Domain	The set of input values for which the function is defined.	For ${\rm{f}}({\rm{x}}) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$, the denominator ${\rm{x}}^2+1$ is never zero for real x. The domain is all real numbers, ${\rm{x}} \in {\rm{R}}$.
Range	The set of output values the function can produce.	The range of ${\rm{f}}({\rm{x}})$ is $[-1, 1)$. The minimum is -1, the maximum is approached but not reached (1).
Critical Point	A point in the domain where the derivative is zero or undefined. Potential locations for local extrema.	${\rm{f}}'({\rm{x}}) = \frac{{4{\rm{x}}}}{{({\rm{x}}^2 + 1)^2}}$. Critical point at x=0 where ${\rm{f}}'(0) = 0$.

Additional Information: Function Analysis

Understanding the function ${\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$ further helps grasp its behavior.

Symmetry: ${\rm{f}}(-{\rm{x}}) = \frac{{({-x})^2} - 1}{({-x})^2 + 1} = \frac{{{\rm{x}}^2} - 1}{{{\rm{x}}^2} + 1} = {\rm{f}}({\rm{x}})$. The function is an even function, symmetric about the y-axis. This confirms that if there is an extremum at $x=0$, it's the only one due to symmetry and the single critical point.
Horizontal Asymptote: As ${\rm{x}} \to \pm \infty$, ${\rm{f}}({\rm{x}}) \to 1$. The line $y=1$ is a horizontal asymptote. This means the function values get closer and closer to 1 as $|{\rm{x}}|$ becomes very large.
Vertical Asymptote: Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. Since ${\rm{x}}^2+1$ is never zero for real x, there are no vertical asymptotes.
Graph: The graph starts approaching y=1 from below as $x \to -\infty$, decreases to its minimum value of -1 at $x=0$, then increases again approaching y=1 from below as $x \to \infty$.

Prepp Team · Answer

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Prepp Team · Answer

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