Problem 16 - Calculus for Biology and Medicine

Question

Let$$f(x)=\frac{2}{1-x^{2}}, \quad x \neq-1,1$$ (a) Show that $$\lim _{x \rightarrow+\infty} f(x)=0$$ and $$\lim _{x \rightarrow-\infty} f(x)=0$$ That is, show that \(y=0\) is a horizontal asymptote of \(f(x)\). (b) Show that $$\lim _{x \rightarrow-1^{-}} f(x)=-\infty$$ and $$\lim _{x \rightarrow-1^{+}} f(x)=+\infty$$ and that $$\lim _{x \rightarrow 1^{-}} f(x)=+\infty$$ and $$\lim _{x \rightarrow 1^{+}} f(x)=-\infty$$ That is, show that \(x=-1\) and \(x=1\) are vertical asymptotes of \(f(x)\) (c) Determine where \(f(x)\) is increasing and where it is decreasing. Does \(f(x)\) have local extrema? (d) Determine where \(f(x)\) is concave up and where it is concave down. Does \(f(x)\) have inflection points? (e) Sketch the graph of \(f(x)\) together with its asymptotes.

Step-by-Step Solution

Answer

The function has horizontal asymptote at \( y=0 \), vertical asymptotes at \( x=-1,1 \), increases on \( x>0 \), decreases on \( x<0 \), concave up near zero, no local extrema or inflection points.

Step 1: Determine Horizontal Asymptotes

To find the horizontal asymptotes, evaluate \[ \lim_{x \to +\infty} f(x) \text{ and } \lim_{x \to -\infty} f(x) \]Given that \[ f(x) = \frac{2}{1-x^2} \]As \( x \to \pm\infty \), \( x^2 \to +\infty \) and hence \( 1-x^2 \to -\infty \). Therefore, the fraction approaches 0.Thus, \[ \lim_{x \to +\infty} f(x) = 0 \text{ and } \lim_{x \to -\infty} f(x) = 0 \]This indicates \( y = 0 \) is a horizontal asymptote.

Step 2: Determine Vertical Asymptotes at x=-1 and x=1

Vertical asymptotes occur where the function becomes unbounded.Observe that:\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{2}{1-x^2} \]Substitute \( x = -1 + h \), where \( h \to 0^- \), making \( 1 - (-1 + h)^2 = 1 - 1 - 2h - h^2 \approx -2h \).Then,\[ \frac{2}{-2h} \to -\infty \] Thus, \( \lim_{x \to -1^-} f(x) = -\infty \).Similarly,For \( x \to -1^+ \), substitute \( x = -1 + h \) where \( h \to 0^+ \):\[ \frac{2}{1-(-1+h)^2} \approx \frac{2}{-2h} \to +\infty \]Thus, \( \lim_{x \to -1^+} f(x) = +\infty \).For \( x \to 1^- \) and \( x \to 1^+ \), similar analysis yields:\[ \lim_{x \to 1^-} f(x) = +\infty \text{ and } \lim_{x \to 1^+} f(x) = -\infty \]Therefore, \( x = -1 \) and \( x = 1 \) are vertical asymptotes.

Step 3: Find Intervals of Increase and Decrease

Differentiate \( f(x) \) to determine where it is increasing or decreasing:\[ f'(x) = \frac{d}{dx} \left( \frac{2}{1-x^2} \right) = \frac{d}{dx} \left( 2(1-x^2)^{-1} \right) \]Using the chain rule:\[ f'(x) = 2(-1)(1-x^2)^{-2}(0 - 2x) = \frac{4x}{(1-x^2)^2} \]Analyze \( f'(x) \):- \( f'(x) > 0 \) for \( x > 0 \) (increasing)- \( f'(x) < 0 \) for \( x < 0 \) (decreasing)Thus, \( f(x) \) is increasing for \( x > 0 \) and decreasing for \( x < 0 \). Since there is no point where \( f'(x)=0 \), there are no local extrema.

Step 4: Determine Concavity and Inflection Points

Find \( f''(x) \) to check for concavity:Take the derivative of \( f'(x) \):\[ f''(x) = \frac{d}{dx} \left( \frac{4x}{(1-x^2)^2} \right) = \frac{4(1-x^2)^2 - 4x \cdot -2x(1-x^2)}{(1-x^2)^4} \]Simplify:\[ f''(x) = \frac{4(1-x^2)(1+x^2) + 16x^2}{(1-x^2)^4} \]Determine the sign of \( f''(x) \):- \( f''(x) > 0 \) where \( x^2 < \frac{1}{3} \) indicates concave up- \( f''(x) < 0 \) elsewhere indicates concave downInflection points occur where \( f''(x) = 0 \), but these only change concavity if \( x \) is in the domain.

Step 5: Sketch f(x) with Asymptotes

Using the information from previous steps:1. Horizontal asymptote at \( y = 0 \).2. Vertical asymptotes at \( x = -1 \) and \( x = 1 \).3. Increasing on \( x > 0 \), decreasing on \( x < 0 \).4. Concave up for \( x^2 < \frac{1}{3} \), concave down elsewhere.Draw the graph with asymptotic behavior and curve characteristics.

Key Concepts

Limits

Limits are a fundamental concept in calculus that describe the behavior of a function as the input approaches a particular value or infinity. In this exercise, we investigate the limits of the function \(f(x) = \frac{2}{1-x^2}\) as \(x\) approaches positive and negative infinity.
As \(x\) goes to infinity, the denominator \(1-x^2\) becomes very large negatively because \(x^2\) grows faster than \(1\). Hence, \(\frac{2}{1-x^2}\) approaches zero since dividing a constant by an ever-increasing magnitude results in a value closer to zero.
Similarly, as \(x\) approaches negative infinity, the same logic applies, leading to the limit being zero. These limits indicate that the horizontal asymptote of the function is \(y = 0\). Understanding limits helps predict the end behavior of functions.

Asymptotes

Asymptotes are lines that a graph approaches but never actually reaches. Knowing the type and position of asymptotes is crucial to graphing functions.
There are two types of asymptotes in this case:

• Horizontal Asymptote: For \(f(x) = \frac{2}{1-x^2}\), the horizontal asymptote is found by considering the limits as \(x\) approaches infinity. Here, \(y = 0\) is the horizontal asymptote.
• Vertical Asymptotes: These occur where the function \(f(x)\) becomes unbounded. For \(x = -1\) and \(x = 1\), \(1-x^2 = 0\), making the function undefined, leading to vertical asymptotes at these points.

Vertical asymptotes suggest the function will shoot towards positive or negative infinity as it gets arbitrarily close to the values \(x = -1\) and \(x = 1\).

Derivative

A derivative represents the instantaneous rate of change of a function or the slope of the tangent line at any point. This concept helps determine where a function is increasing or decreasing.
For \(f(x) = \frac{2}{1-x^2}\), we use the chain rule to find its derivative: \(f'(x) = \frac{4x}{(1-x^2)^2}\). This derivative is positive when \(x > 0\) and negative when \(x < 0\). Thus, the function is increasing when \(x > 0\) and decreasing when \(x < 0\).
No local extrema exist due to the derivative never equalling zero, which would indicate a stationary point.

Concavity

Concavity describes the curvature direction of the graph of a function. A function is called concave up if it curves upwards like a cup, and concave down if it curves downwards.
To find concavity, we calculate the second derivative, \(f''(x)\), of \(f(x) = \frac{2}{1-x^2}\). Here, \(f''(x) = \frac{4(1-x^2)(1+x^2) + 16x^2}{(1-x^2)^4}\).
After simplifying, we note:

• \(f''(x) > 0\) when \(x^2 < \frac{1}{3}\), indicating the function is concave up in this region.
• \(f''(x) < 0\) elsewhere, indicating concave down.

Changes in concavity often reveal inflection points, but we must check if these points are within our domain to confirm their validity.