Problem 42 - Algebra and Trigonometry

Question

\(35-42\) A function is given. (a) Find all the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$ V(x)=\frac{1}{x^{2}+x+1} $$

Step-by-Step Solution

Answer

Local minimum at \(x = -0.5\). Increasing on \((-\

Step 1: Find the Derivative

To find local maxima and minima, we first find the derivative of the function. Given \( V(x) = \frac{1}{x^2+x+1} \), use the quotient rule: \( V'(x) = \frac{-(2x+1)}{(x^2+x+1)^2} \).

Step 2: Set the Derivative to Zero

Set the derivative \( V'(x) \) to zero to find the critical points. Solve \( \frac{-(2x+1)}{(x^2+x+1)^2} = 0 \). This equation simplifies to \( 2x+1 = 0 \), giving the critical point \( x = -0.5 \).

Step 3: Determine Concavity with the Second Derivative

Next, find the second derivative \( V''(x) \) to determine the nature of the critical point. Differentiating \( V'(x) \), we get \( V''(x) = \frac{2x^2 + 4x + 1}{(x^2+x+1)^3} \). Calculate \( V''(-0.5) \) to find out whether it's a maximum or minimum point.

Step 4: Evaluate Second Derivative at Critical Point

Substitute \(-0.5\) into the second derivative: \( V''(-0.5) = \frac{2(-0.5)^2 + 4(-0.5) + 1}{((-0.5)^2-0.5+1)^3} = \frac{0.5}{(0.75)^3} > 0 \). Since \( V''(-0.5) > 0 \), \( x = -0.5 \) is a local minimum.

Step 5: Find Intervals of Increase and Decrease

To determine intervals of increase or decrease, analyze the sign of \( V'(x) \). Since \( V'(x) = \frac{-(2x+1)}{(x^2+x+1)^2} \), \( V(x) \) is increasing when \( V'(x) > 0 \) and decreasing when \( V'(x) < 0 \). Since \( 2x+1\) changes sign at \( x = -0.5 \), we find that \( V(x) \) is increasing for \( x < -0.5 \) and decreasing for \( x > -0.5 \).

Key Concepts

Local Maximum and Minimum

In calculus, finding the local maximum and minimum involves identifying where a function changes its direction of growth. A local maximum is where the function attains its highest value in a certain range, while a local minimum is the lowest. These points provide crucial insights into the behavior of the function.

To find these points, we start by finding the derivative of the function, which helps us determine where the slope of the tangent is zero or undefined, indicating a potential maximum or minimum. As shown in the solution, we use the result of setting the first derivative to zero to locate the critical points.

In the given problem with the function \( V(x) = \frac{1}{x^2+x+1} \), we found \( x = -0.5 \) to be a critical point. By using the second derivative test, we confirmed it's a local minimum because the second derivative was positive at this point, indicating the function is concave up at \( x = -0.5 \).

Derivative

The derivative is a fundamental concept in calculus. It measures the rate at which a function is changing at any given point. Geometrically, it represents the slope of the tangent line to the curve at that point.

For the function \( V(x) = \frac{1}{x^2+x+1} \), the derivative \( V'(x) \) was found using the quotient rule, a technique used because the function is in the form of one expression divided by another. The quotient rule is expressed as:

• If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

By applying this, we found \( V'(x) = \frac{-(2x+1)}{(x^2+x+1)^2} \). Setting this equal to zero allowed us to solve for the critical points, crucial in determining where the function might have local maxima or minima.

In this solution, locating the critical points was an essential step before applying the second derivative test to differentiate between local maxima and minima.

Concavity

Concavity describes the direction a function curves. A function is concave up when its slope increases, forming a U-shape, and concave down when the slope decreases, forming an upside-down U-shape. These characteristics are determined using the second derivative.

The second derivative test helps in assessing concavity:

• If the second derivative is positive, the function is concave up at that point, suggesting a potential local minimum.
• If the second derivative is negative, the function is concave down, indicating a potential local maximum.

In the solved exercise, when we computed the second derivative: \( V''(x) = \frac{2x^2 + 4x + 1}{(x^2+x+1)^3} \) and evaluated it at \( x = -0.5 \), we found it to be positive. This confirmed the function was concave up, thus solidifying our conclusion that \( x = -0.5 \) is a local minimum.

Understanding concavity is important for sketching functions and delivering insights into the nature of critical points we found with the first derivative.