Problem 30 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=\frac{x}{\sqrt{x^{2}-1}} $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x}{\sqrt{x^2-1}}\) is increasing on intervals \((- \infty, -1)\) and \((1, \infty)\) and decreasing on interval \((-1, 1)\). There are no relative maxima or minima for this function, as it is undefined at the critical points \(x=\pm1\).

Step 1: Find the derivative of the function

We start by finding the derivative of \(f(x)\) using the quotient rule. If \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x)=\frac{u'(x)v(x) - v'(x)u(x)}{v^2(x)}\). In our case, \(u(x)=x\) and \(v(x)=\sqrt{x^2-1}\). Thus, we have: \[u'(x) = 1\] \[v'(x) = \frac{d}{dx} (\sqrt{x^2 - 1}) = \frac{d}{dx} ((x^2 - 1)^{\frac{1}{2}})\] To find \(v'(x)\), we apply the chain rule: \[v'(x) = \frac{1}{2}(x^2 - 1)^{-\frac{1}{2}}\cdot (2x) = \frac{x}{\sqrt{x^2-1}}\] Now, we can apply the quotient rule and find \(f'(x)\): \[f'(x) = \frac{1 \cdot \sqrt{x^2 - 1} - x \cdot \frac{x}{\sqrt{x^2 - 1}}}{(\sqrt{x^2 - 1})^2}\] \[f'(x) = \frac{x^2 - (x^2 - 1)}{x^2 - 1}\] \[f'(x) = \frac{1}{x^2-1}\]

Step 2: Determine critical points

Now, we find the critical points where the derivative \(f'(x)\) is either equal to zero or undefined. In this case, the derivative is never equal to zero, but it's undefined when the denominator equals zero: \[x^2 - 1 = 0\] \[x^2 = 1\] \[x = \pm 1\] Now, we have two critical points, \(x = 1\) and \(x = -1\).

Step 3: Analyze intervals and determine if f(x) is increasing or decreasing

To find out whether the function is increasing or decreasing in the intervals determined by the critical points, we evaluate the derivative at points inside these intervals. The intervals are \((- \infty, -1)\), \((-1, 1)\), and \((1, \infty)\). For \(x = -2\), \(f'(-2) = \frac{1}{(-2)^2-1} = \frac{1}{3} > 0\), so \(f(x)\) is increasing on the interval \((- \infty, -1)\). For \(x = 0\), \(f'(0) = \frac{1}{0^2-1} = -1 < 0\), so \(f(x)\) is decreasing on the interval \((-1, 1)\). For \(x = 2\), \(f'(2) = \frac{1}{(2)^2-1} = \frac{1}{3} > 0\), so \(f(x)\) is increasing on the interval \((1, \infty)\).

Step 4: Identify relative maxima and minima

Based on the analysis of intervals, we can infer that there is a relative minimum at \(x = -1\) and a relative maximum at \(x = 1\). We find the corresponding values of the function: \(f(-1) = \frac{-1}{\sqrt{(-1)^2 - 1}}\) is undefined. \(f(1) = \frac{1}{\sqrt{(1)^2 - 1}}\) is undefined. However, as the function is undefined at these points, there are no relative maxima or minima for this function.

Key Concepts

Quotient Rule

When differentiating a function that is the ratio of two other functions, we use the quotient rule. This rule states that the derivative of a function expressed as \( f(x) = \frac{u(x)}{v(x)} \) is \( f'(x) = \frac{u'(x)v(x) - v'(x)u(x)}{v^2(x)} \). The process involves finding the derivatives of the numerator \( u(x) \) and the denominator \( v(x) \) separately before applying the rule.

For a function like \( f(x) = \frac{x}{\sqrt{x^{2}-1}} \), we identify \( u(x) = x \) and thus, its derivative is \( u'(x) = 1 \). The denominator, \( v(x) = \sqrt{x^2-1} \), is a composite function which requires an additional differentiation technique, the chain rule, to find its derivative \( v'(x) \). Once we have both derivatives, we apply the quotient rule to find \( f'(x) \) which gives us insight on the function's increasing and decreasing intervals.

Chain Rule

The chain rule is a powerful tool for differentiating compositions of functions. When we have a function within a function, like \( v(x)= \sqrt{x^2-1} \) which can be rewritten as \( (x^2-1)^{\frac{1}{2}} \), the chain rule allows us to handle the inner function separately.

For our example, we differentiate the inner function \( x^2-1 \) and the outer function \( u^{\frac{1}{2}} \) where, in this case, \( u = x^2-1 \). The chain rule formula is expressed as \( \frac{d}{dx} \left[g(h(x))\right] = g'(h(x)) \cdot h'(x) \). Applying the chain rule to \( v(x) \) gives us \( v'(x) = \frac{x}{\sqrt{x^2-1}} \). It's essential to grasp the chain rule to differentiate functions like our \( f(x) \) properly, especially when using the quotient rule.

Critical Points

The concept of critical points is fundamental when analyzing the behavior of a function. A critical point exists where the derivative of the function is either zero or undefined. These points are potential locations of relative maxima, minima, or inflection points.

To find the critical points for \( f(x) = \frac{x}{\sqrt{x^{2}-1}} \), we set the derivative \( f'(x) = \frac{1}{x^2-1} \) equal to zero and also look for where it might be undefined. Since the derivative cannot be zero, the critical points come from the values that make the derivative undefined, which are at \( x = \pm 1 \). Even though these are classified as critical points, it is crucial to remember that they might not correspond to actual points on the function where relative maxima or minima occur.

Relative Maxima and Minima

The terms relative maxima and minima refer to the peaks and valleys of a function when looking at specific intervals. These can be found using the first derivative test, which involves looking at the sign of the derivative before and after a critical point. A change from positive to negative indicates a relative maximum, while a change from negative to positive suggests a relative minimum.

In our problem, we are led to believe that there may be a relative maximum at \( x = 1 \) and a relative minimum at \( x = -1 \) because of the function's behavior around these points. However, since \( f(\pm 1) \) is undefined, these critical points do not correspond to relative maxima or minima on the graph of the function. It's important to check both the critical points and the actual function values to determine the presence of relative extrema accurately.