Key Concepts and Formulas

• Local Minima/Maxima: A local minimum (or maximum) of a function occurs at a point where the function's value is smaller (or larger) than all nearby points.
• First Derivative Test: Critical points (where the first derivative is zero or undefined) are potential locations of local extrema.
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $x = c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $x = c$.

Step-by-Step Solution

Step 1: Define the functions and $h(x)$

We are given the functions: $f\left( x \right) = {x^2} + {1 \over {{x^2}}}$ $g\left( x \right) = x - {1 \over x}$ And $h(x)$ is defined as their ratio: $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$ The domain for $x$ is $R - \left\{ { - 1,0,1} \right\}$. This ensures that $g(x)$ is not zero (as $x \ne \pm 1$) and the terms $1/x^2$ and $1/x$ are well-defined (as $x \ne 0$).

Step 2: Simplify $h(x)$ using algebraic manipulation

Substitute the expressions for $f(x)$ and $g(x)$ into $h(x)$: $h\left( x \right) = {{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$ Our goal is to simplify this expression. Notice that the denominator is $x - \frac{1}{x}$. Let's try to express the numerator, $x^2 + \frac{1}{x^2}$, in terms of $x - \frac{1}{x}$. Recall the algebraic identity: $(a-b)^2 = a^2 - 2ab + b^2$. Let $a=x$ and $b=\frac{1}{x}$. Then: ${\left( {x - {1 \over x}} \right)^2} = {x^2} - 2 \cdot x \cdot {1 \over x} + {1 \over {{x^2}}}$ ${\left( {x - {1 \over x}} \right)^2} = {x^2} - 2 + {1 \over {{x^2}}}$ From this, we can express the numerator: ${x^2} + {1 \over {{x^2}}} = {\left( {x - {1 \over x}} \right)^2} + 2$ Now, substitute this back into the expression for $h(x)$: $h\left( x \right) = {{{{\left( {x - {1 \over x}} \right)}^2} + 2} \over {x - {1 \over x}}}$ Why this step? This algebraic manipulation is crucial because it allows us to express $h(x)$ entirely in terms of the common expression $x - \frac{1}{x}$. This significantly simplifies the function and prepares it for a substitution.

Step 3: Introduce a substitution to simplify the function further

Let $t = x - {1 \over x}$. Now, $h(x)$ can be rewritten as a function of $t$, let's call it $H(t)$: $H\left( t \right) = {{{t^2} + 2} \over t}$ This can be further simplified by dividing each term in the numerator by $t$: $H\left( t \right) = {t^2 \over t} + {2 \over t} = t + {2 \over t}$ Why this step? By substituting $t$, we transform a function of $x$ into a much simpler function of $t$. Differentiating $t + \frac{2}{t}$ is far easier than differentiating the original $h(x)$ using the quotient rule.

Step 4: Find the critical points using the First Derivative Test

To find the local minimum or maximum values of $H(t)$, we need to find its critical points by differentiating $H(t)$ with respect to $t$ and setting the derivative to zero. $H'\left( t \right) = {d \over {dt}}\left( {t + {2 \over t}} \right)$ $H'\left( t \right) = {d \over {dt}}\left( t \right) + {d \over {dt}}\left( {2{t^{ - 1}}} \right)$ $H'\left( t \right) = 1 + 2\left( { - 1} \right){t^{ - 2}}$ $H'\left( t \right) = 1 - {2 \over {{t^2}}}$ Now, set $H'(t) = 0$ to find the critical points: $1 - {2 \over {{t^2}}} = 0$ ${1 = {2 \over {{t^2}}}}$ ${{t^2} = 2}$ ${t = \pm \sqrt 2 }$ So, our critical points are $t = \sqrt{2}$ and $t = -\sqrt{2}$. Why this step? Setting the first derivative to zero helps us locate points where the slope of the tangent to the function is horizontal. These are potential points of local extrema.

Step 5: Apply the Second Derivative Test to classify the critical points

To determine whether these critical points correspond to a local minimum or maximum, we use the Second Derivative Test. We need to calculate $H''(t)$. $H''\left( t \right) = {d \over {dt}}\left( {1 - {2 \over {{t^2}}}} \right)$ $H''\left( t \right) = {d \over {dt}}\left( 1 \right) - {d \over {dt}}\left( {2{t^{ - 2}}} \right)$ $H''\left( t \right) = 0 - 2\left( { - 2} \right){t^{ - 3}}$ $H''\left( t \right) = {4 \over {{t^3}}}$ Now, evaluate $H''(t)$ at the critical points:

• For $t = \sqrt{2}$: $H''\left( {\sqrt 2 } \right) = {4 \over {{{\left( {\sqrt 2 } \right)}^3}}} = {4 \over {2\sqrt 2 }} = {2 \over {\sqrt 2 }} = \sqrt 2 > 0$ Since $H''(\sqrt{2}) > 0$, $H(t)$ has a local minimum at $t = \sqrt{2}$.
• For $t = -\sqrt{2}$: $H''\left( { - \sqrt 2 } \right) = {4 \over {{{\left( { - \sqrt 2 } \right)}^3}}} = {4 \over { - 2\sqrt 2 }} = { - 2 \over {\sqrt 2 }} = - \sqrt 2 < 0$ Since $H''(-\sqrt{2}) < 0$, $H(t)$ has a local maximum at $t = -\sqrt{2}$. Why this step? The sign of the second derivative at a critical point tells us about the concavity of the function, which helps us classify the critical point as a local minimum or maximum.

Step 6: Calculate the local minimum value

We found that $H(t)$ has a local minimum at $t = \sqrt{2}$. To find the local minimum value, substitute $t = \sqrt{2}$ back into the expression for $H(t)$: $H\left( {\sqrt 2 } \right) = \sqrt 2 + {2 \over {\sqrt 2 }} = \sqrt 2 + {{\sqrt 2 \cdot \sqrt 2 } \over {\sqrt 2 }} = \sqrt 2 + \sqrt 2 = 2\sqrt 2$ Why this step? Once we've identified the point at which the local minimum occurs, we evaluate the function at that point to find the actual minimum value.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when simplifying expressions, especially when dealing with fractions and exponents.
• Domain Restrictions: Always consider the domain of the original function when performing substitutions. Ensure that the values of the new variable are valid.
• Second Derivative Test Confusion: Remember the conditions for the Second Derivative Test: $H''(t) > 0$ implies a local minimum, and $H''(t) < 0$ implies a local maximum.

Summary

We found the local minimum value of $h(x)$ by first simplifying the expression using algebraic manipulation and a substitution. This transformed the problem into finding the local minimum of a simpler function $H(t)$. We then used the first and second derivative tests to find the critical points and classify them. Finally, we evaluated $H(t)$ at the local minimum to obtain the local minimum value of $2\sqrt{2}$.

Final Answer The final answer is \boxed{2\sqrt 2}, which corresponds to option (A).