Problem 2 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ f(x)=x^{2}-6 x+1 $$

Step-by-Step Solution

Answer

Function decreases on \((-\text{∞}, 3)\) and increases on \((3, \text{∞})\). Concave up on \((-\text{∞}, \text{∞})\). Local minimum at (3, -8).

Step 1: Find the First Derivative of the Function

To determine the intervals of increase and decrease, find the first derivative of the function. Given function: \[ f(x) = x^2 - 6x + 1 \]. First derivative: \[ f'(x) = \frac{d}{dx}(x^2 - 6x + 1) = 2x - 6 \]

Step 2: Find Critical Points

Set the first derivative equal to zero and solve for x. \[ 2x - 6 = 0 \]. Solving this gives \[ x = 3 \].

Step 3: Determine Intervals of Increase and Decrease

Determine the sign of the first derivative on each interval determined by the critical points. For \( x < 3 \): Choose \( x = 2 \), \[ f'(2) = 2(2) - 6 = -2 \], (negative means decreasing). For \( x > 3 \): Choose \( x = 4 \), \[ f'(4) = 2(4) - 6 = 2 \], (positive means increasing). So, the function is decreasing on the interval \((-\text{∞}, 3)\) and increasing on the interval \((3, \text{∞})\).

Step 4: Find the Second Derivative of the Function

To determine intervals of concavity, find the second derivative of the function. Second derivative: \[ f''(x) = \frac{d}{dx}(2x - 6) = 2 \].

Step 5: Determine Intervals of Concavity

Since the second derivative \( f''(x) = 2 \) is always positive, the function is concave up on the entire domain \((-\text{∞}, \text{∞})\). There are no inflection points.

Step 6: Find Intercepts

To find the y-intercept, set \( x = 0 \) in the original function. \[ f(0) = 0^2 - 6(0) + 1 = 1 \]. So, the y-intercept is \( (0, 1) \). To find the x-intercept(s), set \( f(x) = 0 \). Solve \( x^2 - 6x + 1 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = -6, c = 1 \), the x-intercepts are \( x = 3 \pm 2\sqrt{2} \) (approximately 0.586 and 5.414).

Step 7: Identify High and Low Points

The first derivative indicates that there is a local minimum at \( x = 3 \). Evaluate the function at this point: \[ f(3) = 3^2 - 6(3) + 1 = -8 \]. So, the local minimum point is \( (3, -8) \).

Step 8: Sketch the Graph

Sketch the function using the information obtained: - Decreasing on \((-\text{∞}, 3)\). - Increasing on \((3, \text{∞})\). - Concave up on \((-\text{∞}, \text{∞})\). - y-intercept at \( (0, 1) \). - x-intercepts at approximately \( (0.586, 0) \) and \( (5.414, 0) \). - Local minimum at \( (3, -8) \).

Key Concepts

derivatives

In calculus, derivatives are fundamental. They measure the rate at which a function changes. If you think of a function as a curve, the derivative at a particular point gives you the slope of the tangent line to the curve at that point.

For a function \( f(x) \), its first derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). The first derivative helps us identify critical points, where the slope is zero, indicating potential local maxima or minima.

intervals of increase and decrease

To find the intervals where a function is increasing or decreasing, we use its first derivative. When \( f'(x) > 0 \), the function \( f(x) \) is increasing, meaning the curve slopes upwards. Conversely, when \( f'(x) < 0 \), the function is decreasing, and the curve slopes downwards.

We check the sign of the first derivative in different intervals. These intervals are determined by the critical points. If a function changes from positive to negative at a critical point, it suggests a local maximum. If it changes from negative to positive, it indicates a local minimum.

concavity

Concavity refers to the direction in which a function curves. When we talk about a function being 'concave up' or 'concave down', we are describing its curvature.

The second derivative of the function, denoted as \( f''(x) \), helps us determine concavity. When \( f''(x) > 0 \) at all points in an interval, the function is concave up (like a cup). This means that the function curves upwards and resembles an upward-facing bowl. On the other hand, if \( f''(x) < 0 \), the function is concave down (like a frown), curving downwards like an upside-down bowl.

Points where the concavity changes from concave up to concave down, or vice versa, are called points of inflection.