Problem 27 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x-\ln (x) $$

Step-by-Step Solution

Answer

The function is concave up on (0, ∞) with a local minimum at x = 1.

Step 1: Find the First Derivative

First, we need to calculate the first derivative of the function to find the critical points. The given function is \( f(x) = x - \ln(x) \). Applying basic differentiation rules, we get:\[ f'(x) = 1 - \frac{1}{x} \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points:\[ 1 - \frac{1}{x} = 0 \]Solving this equation, we find:\[ 1 = \frac{1}{x} \] which implies \( x = 1 \) is a critical point.

Step 3: Find the Second Derivative

Next, we calculate the second derivative to determine concavity and points of inflection. Differentiate the first derivative \( f'(x) = 1 - \frac{1}{x} \):\[ f''(x) = \frac{1}{x^2} \]

Step 4: Determine Concavity and Point of Inflection

The second derivative \( f''(x) = \frac{1}{x^2} \) is always positive for \( x > 0 \). Therefore:- The function is concave up on \( (0, \infty) \).- There are no intervals where the function is concave down.- Since \( f''(x) \) does not change sign, there are no points of inflection.

Step 5: Use the Second Derivative Test for Critical Point

Since \( f''(1) = \frac{1}{1^2} > 0 \), the critical point \( x = 1 \) is a point of local minimum by the Second Derivative Test. There are no other critical points.

Key Concepts

Second Derivative Test

The Second Derivative Test is a handy tool in calculus that allows you to classify critical points as local maximums or minimums by examining the value of the second derivative at those points. To perform this test:

• First, find the critical points by setting the first derivative of the function equal to zero and solving for the variable.
• Next, compute the second derivative of the function.
• To apply the test, evaluate the second derivative at each critical point:
• If the second derivative is positive \(f''(c) > 0\) at the critical point \(x = c\), the function has a local minimum there.
• If the second derivative is negative \(f''(c) < 0\), then it's a local maximum.
• If the second derivative equals zero \(f''(c) = 0\), the test is inconclusive.

For the function \(f(x) = x - \ln(x)\), we calculated that at the critical point \(x = 1\), the second derivative is positive. Hence, \(x = 1\) is a local minimum.

Concave Up and Concave Down Intervals

Finding concave up and concave down intervals involves seeing where the function curves upwards or downwards. This characteristic is determined by the sign of the second derivative.

• When the second derivative \(f''(x)\) is positive, the graph of the function is concave up. Imagine a U-shaped curve."
• If the second derivative is negative, the function is concave down, similar to an upside-down U.

For our function \(f(x) = x - \ln(x)\), the second derivative \(f''(x) = \frac{1}{x^2}\) is always positive for \(x > 0\). This means the function is concave up on the interval \( (0, \infty)\). Since there are no negative values, the function is never concave down.

Points of Inflection

Points of inflection occur where the function changes concavity, shifting from concave up to concave down or vice versa. This is typically marked by a change in the sign of the second derivative.

• A point of inflection is a location on the graph where the tangent line crosses the graph itself.
• Mathematically, you find these points by setting the second derivative equal to zero or undefined, then checking changes in sign around those points.

In our exercise, the second derivative \(f''(x) = \frac{1}{x^2}\) never changes its sign because it is always positive for \(x > 0\). Hence, \(f(x) = x - \ln(x)\) does not have any points of inflection.

Critical Points Analysis

Critical points are where the first derivative of the function is zero or undefined. These points are potential locations for local maxima, minima, or points of inflection.

• To find critical points, solve the equation \(f'(x) = 0\) for \(x\). If the derivative does not exist, it may also be a critical point.
• Once you have potential points, use tests like the Second Derivative Test to classify them further.

For \(f(x) = x - \ln(x)\), the first derivative \(f'(x) = 1 - \frac{1}{x}\) was set to zero, yielding \(x = 1\) as the critical point. The Second Derivative Test confirmed that this is a point of local minimum due to the positive second derivative value at this point. No other critical points exist for this function.