Problem 229 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f \text { , }} \\\ {\text { c. intervals where } f \text { is concave up and concave }} \\\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . }}\end{array} $$ $$ f(x)=x^{2}+x+1 $$

Step-by-Step Solution

Answer

f is decreasing on \((-\infty, -\frac{1}{2})\), increasing on \((-\frac{1}{2}, \infty)\); local minimum at \( x = -\frac{1}{2} \), no local maximum; concave up everywhere, no inflection points.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we start by finding the first derivative of the function \( f(x) = x^2 + x + 1 \). The first derivative is \( f'(x) = 2x + 1 \).

Step 2: Identify Critical Points

The function is increasing where the first derivative is positive, and decreasing where it's negative. Set the derivative equal to zero to find critical points: \( 2x + 1 = 0 \). Solving for \( x \), we get \( x = -\frac{1}{2} \).

Step 3: Analyze Intervals for Increasing or Decreasing Behavior

(a) Test the intervals around the critical point \( x = -\frac{1}{2} \). For \( x < -\frac{1}{2} \), pick \( x = -1 \): \( f'(-1) = 2(-1) + 1 = -1 \), so decreasing. For \( x > -\frac{1}{2} \), pick \( x = 0 \): \( f'(0) = 2(0) + 1 = 1 \), so increasing. Therefore, \( f \) is decreasing on \((-\infty, -\frac{1}{2})\) and increasing on \((-\frac{1}{2}, \infty)\).

Step 4: Find Critical Values and Determine Local Extrema

(b) Since \( f \) switches from decreasing to increasing at \( x = -\frac{1}{2} \), this point is a local minimum. The local minimum value of the function is \( f(-\frac{1}{2}) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} \). There is no local maximum since the function never switches from increasing to decreasing.

Step 5: Find the Second Derivative

To analyze concavity, we find the second derivative: \( f''(x) = 2 \).

Step 6: Determine Concavity and Inflection Points

(c and d) Since \( f''(x) = 2 \) is positive for all \( x \), the function is concave up everywhere and concave down nowhere, so there are no inflection points.

Key Concepts

Increasing and Decreasing Functions

When working with functions, you'll often want to determine where a function is increasing or decreasing. This tells you where the function's slope is positive or negative. The

• First, calculate the first derivative of the function. For our exercise with the function \( f(x) = x^2 + x + 1 \), this first derivative is \( f'(x) = 2x + 1 \).
• The function increases when \( f'(x) > 0 \) and decreases when \( f'(x) < 0 \).
• To find these intervals, set \( f'(x) = 0 \) to find the critical points where the function changes from increasing to decreasing or vice versa.

For the given function, solving \( 2x + 1 = 0 \) gives the critical point \( x = -\frac{1}{2} \). Evaluating the derivative around this point helps identify intervals:

• For \( x < -\frac{1}{2} \): The function decreases.
• For \( x > -\frac{1}{2} \): The function increases.

This is a foundational tool in calculus for sketching graphs and understanding function behavior.

Critical Points

Critical points of a function are values of \( x \) where the first derivative \( f'(x) \) is zero or undefined. They are essential in determining potential local maxima or minima. For the function \( f(x) = x^2 + x + 1 \):

• Set the first derivative \( f'(x) = 2x + 1 \) equal to zero.
• Solving \( 2x + 1 = 0 \) gives the critical point \( x = -\frac{1}{2} \).

This critical point identifies where the function could have a local extrema.
Once identified, further testing is required to determine whether these points are maxima, minima or neither by checking if the function changes direction at these points. Here, the function decreases before and increases after \( x = -\frac{1}{2} \), indicating a local minimum at \( x = -\frac{1}{2} \). Evaluating \( f(-\frac{1}{2}) \) gives us a local minimum value of \( \frac{3}{4} \). Understanding where these changes occur is crucial for analyzing a function's characteristics.

Concavity

Concavity refers to the direction in which a curve bends. In calculus, we use the second derivative to determine concavity.

• The function is concave up where the second derivative \( f''(x) > 0 \).
• The function is concave down where \( f''(x) < 0 \).

For our function \( f(x) = x^2 + x + 1 \), finding the second derivative gives \( f''(x) = 2 \). This derivative of 2 signifies that the function is always concave up since 2 is positive for all \( x \). Therefore, there are no intervals where the function is concave down.
A function that is concave up bends upwards like a cup and has no flex or turning points since it doesn't change from concave up to concave down. This curvature information helps in understanding the behavior of graphs over intervals.

Second Derivative Test

The second derivative test provides insight on the nature of the critical points you've found. This test can tell you whether a critical point is a local maximum, a local minimum, or neither:

• If \( f''(x) > 0 \) at a critical point, the function has a local minimum there.
• If \( f''(x) < 0 \), the function has a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

In our example with \( f(x) = x^2 + x + 1 \), since \( f''(x) = 2 \) is greater than zero everywhere, the critical point \( x = -\frac{1}{2} \) corresponds to a local minimum. The positive second derivative confirms the curve is opening upwards at that point.
This test is useful when confirming the nature of critical points, especially when graphing or analyzing functional behavior. Knowledge of the second derivative allows mathematicians to understand whether functions curve toward maxima or minima, adding depth to the function analysis.