Problem 235 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. \(f(x)=\frac{1}{1-x}, x \neq 1\)

Step-by-Step Solution

Answer

a. Increasing on \((-\infty, 1)\) and \((1, \infty)\). No local extrema. c. Concave up on \((-\infty, 1)\), concave down on \((1, \infty)\). No inflection points.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the first derivative of the function. Given \( f(x) = \frac{1}{1-x} \), we use the quotient rule for differentiation:\[ f'(x) = \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{0(1-x) - 1(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} \].

Step 2: Determine Intervals of Increase and Decrease

Analyze the sign of \( f'(x) = \frac{1}{(1-x)^2} \). The expression is always positive for all \( x eq 1 \). Therefore, the function \( f(x) \) is increasing on the intervals \((-\infty, 1)\) and \((1, \infty)\).

Step 3: Identify Local Extrema

Since \( f'(x) = \frac{1}{(1-x)^2} \) is never zero or undefined for any \( x eq 1 \), there are no critical points where \( f(x) \) changes from increasing to decreasing. Thus, there are no local minima or maxima.

Step 4: Find the Second Derivative

To analyze concavity, we need the second derivative. From \( f'(x) = \frac{1}{(1-x)^2} \), differentiate again using the chain rule:\[ f''(x) = \frac{d}{dx}\left( \frac{1}{(1-x)^2} \right) = \frac{2}{(1-x)^3} \].

Step 5: Determine Intervals of Concavity

Evaluate the sign of \( f''(x) = \frac{2}{(1-x)^3} \). It is positive on the interval \( (-\infty, 1) \), indicating that \( f(x) \) is concave up there, and negative on \( (1, \infty) \), indicating that \( f(x) \) is concave down.

Step 6: Identify Points of Inflection

An inflection point occurs where the concavity changes. \( f''(x) \) changes sign at \( x=1 \), but since \( f(x) \) is undefined at \( x=1 \), there are no inflection points.

Key Concepts

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing is crucial in analyzing its behavior. For the function \( f(x) = \frac{1}{1-x} \), we first find its first derivative, which tells us the slope of the tangent to the curve at any given point. By using the quotient rule, we derive the first derivative: \[ f'(x) = \frac{1}{(1-x)^2} \] This result is always positive, except where the function isn't defined, at \( x = 1 \). Therefore,

• The function is increasing on the intervals \((-\infty, 1)\) and \((1, \infty)\).

It's important to observe that the denominator \((1-x)^2\) is always positive when \( x eq 1 \), reinforcing that \( f(x) \) increases consistently across these intervals.

Local Minima and Maxima

Local minima and maxima occur where a function changes its direction from increasing to decreasing or vice versa. These points are found at the critical points where the first derivative \( f'(x) \) is zero or undefined. For our function, \( f'(x) = \frac{1}{(1-x)^2} \), which is never zero and only undefined at \( x = 1 \). Therefore,

• There are no points where \( f(x) \) switches its increasing/decreasing behavior.
• Consequently, there are no local minima or maxima for this function.

This behavior is typical for functions like \( f(x) \), which resemble a hyperbola, consistently increasing over its defined range without local peaks or valleys.

Concavity

Examination of a function's concavity reveals its curvature, indicating whether it bends upwards or downwards. The second derivative \( f''(x) \) of \( f(x) \) provides this information. Calculating it, we find: \[ f''(x) = \frac{2}{(1-x)^3} \] To determine concavity:

• If \( f''(x) > 0 \), the function is concave up (curving upwards).
• If \( f''(x) < 0 \), it is concave down (curving downwards).

Evaluating this for our function:

• On \((-\infty, 1)\), \( f''(x) \) is positive meaning it is concave up.
• On \((1, \infty)\), \( f''(x) \) becomes negative, hence it is concave down.

Such analysis can help predict how the function's graph would look, identifying regions of upward and downward bends.

Inflection Points

Inflection points are where a function changes concavity, from concave up to concave down, or vice versa. They occur where the second derivative \( f''(x) \) changes its sign. For the function given: \[ f''(x) = \frac{2}{(1-x)^3} \] The potential inflection point is at \( x = 1 \), where \( f''(x) \) changes from positive (concave up) to negative (concave down). However, the function itself, \( f(x) = \frac{1}{1-x} \), is not defined at \( x = 1 \).

• As \( f(x) \) does not exist at this point, there is actually no inflection point.

Even though there seems to be a sign change, an actual inflection point must lie within a well-defined region of the function. This explains why especially rational functions should be checked for undefined values when seeking inflection points.