Problem 1 - Calculus Single Variable

Question

In each of Exercises \(1-36,\) 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. $$ f(x)=x-4 \sqrt{x} $$

Step-by-Step Solution

Answer

The function is concave up on \((0, \infty)\), has a local minimum at \(x = 4\), and no inflection points.

Step 1: Find the First Derivative

First, we need to find the first derivative of the function \(f(x)=x - 4\sqrt{x}\). To accomplish this, recall that the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). Hence, the first derivative is: \[f'(x) = \frac{d}{dx}(x) - 4\frac{d}{dx}(\sqrt{x}) = 1 - 4 \cdot \frac{1}{2\sqrt{x}} = 1 - \frac{2}{\sqrt{x}}.\]

Step 2: Find the Second Derivative

Next, find the second derivative \(f''(x)\) using the first derivative. Differentiate \(f'(x) = 1 - \frac{2}{\sqrt{x}}\) with respect to \(x\): The derivative of \(-\frac{2}{\sqrt{x}}\) is \(-2 \cdot \frac{d}{dx}(x^{-1/2}) = -2 \cdot (-\frac{1}{2}x^{-3/2}) = \frac{1}{x^{3/2}}\). Thus, \[f''(x) = \frac{d}{dx}(1) + \frac{d}{dx}\left(-\frac{2}{\sqrt{x}}\right) = 0 + \frac{1}{x^{3/2}} = \frac{1}{x^{3/2}}.\]

Step 3: Determine Concavity

The second derivative \(f''(x) = \frac{1}{x^{3/2}}\) tells us about the concavity of \(f(x)\). - Since \(f''(x) > 0\) for all \(x > 0\), \(f(x)\) is concave up on \((0, \infty)\).- There are no intervals where \(f''(x) < 0\), so \(f(x)\) isn't concave down anywhere.

Step 4: Find Points of Inflection

A function has a point of inflection where the sign of \(f''(x)\) changes. Since \(f''(x) = \frac{1}{x^{3/2}}\) is positive and does not change sign, there are no inflection points.

Step 5: Find Critical Points

Critical points occur where \(f'(x) = 0\) or \(f'(x)\) is undefined. Solve \(1 - \frac{2}{\sqrt{x}} = 0\):\[1 = \frac{2}{\sqrt{x}} \, \Rightarrow \, \sqrt{x} = 2 \, \Rightarrow \, x = 4.\] The derivative \(f'(x)\) is undefined at \(x = 0\) due to division by zero.

Step 6: Apply the Second Derivative Test

To determine if the critical point \(x = 4\) is a local minimum or maximum, use the second derivative test. - Compute \(f''(4) = \frac{1}{4^{3/2}} = \frac{1}{8}\).- Since \(f''(4) > 0\), \(x = 4\) is a point of local minimum. There are no local maxima because \(f(x)\) is increasing everywhere else in its domain.

Key Concepts

Concavity

Concavity is concerned with the direction that a curve bends over an interval. For a function like our given function, we can analyze concavity by looking at its second derivative. The sign of the second derivative, denoted as \(f''(x)\), tells us about the concavity:

• If \(f''(x) > 0\), the graph of \(f(x)\) is concave up, resembling a bowl facing upward.
• If \(f''(x) < 0\), the graph of \(f(x)\) is concave down, similar to a bowl facing downward.

In the original exercise, we calculated \(f''(x) = \frac{1}{x^{3/2}}\). Since \(f''(x)\) is positive for every \(x > 0\), the function is concave up across the entire interval \((0, \infty)\). This means our function always curves upwards, so there are no intervals where it bends downwards.

Critical Points

Critical points are the values of \(x\) where the slope of the function is zero or undefined. To find critical points, we need the first derivative, \(f'(x)\), which represents the slope of the tangent to the curve at any point. For the function \(f(x) = x - 4\sqrt{x}\), we derived \(f'(x) = 1 - \frac{2}{\sqrt{x}}\).The critical points occur where this first derivative equals zero or does not exist. - To solve \(1 - \frac{2}{\sqrt{x}} = 0\) results in \(x = 4\), our main critical point in this case.- We also noticed \(f'(x)\) is undefined for \(x = 0\).It is important as these points can potentially be where local minima or maxima are found. Let's use the Second Derivative Test to understand their nature further.

Second Derivative Test

The Second Derivative Test helps in classifying the critical points we just found. It works by examining the second derivative at identified critical points:

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

For our specific function, at critical point \(x = 4\), we computed \(f''(4) = \frac{1}{8}\). Since \(f''(4)\) is positive, it indicates that \(x = 4\) is a point of local minimum. This means around \(x = 4\), the curve dips to its lowest before rising again. The test reaffirms that there are no local maxima in this scenario as \(f(x)\) is always increasing elsewhere.

Inflection Points

Inflection points are where the function changes its curvature direction. It's where \(f''(x)\) changes sign. These points show where functions transition between being concave up or concave down. We were tasked to find these points for \(f(x) = x - 4\sqrt{x}\) by looking at \(f''(x) = \frac{1}{x^{3/2}}\).For this specific function, since \(f''(x)\) remains positive everywhere \(x > 0\), it never changes sign. Therefore, there are no inflection points. This consistent concavity means the curve continues bending in the same way throughout its domain, further reaffirming the absence of transitions to concave down.