Problem 92 - Thomas Calculus

Question

Does the graph of \(y=\sqrt{x}-\ln x, x>0,\) have an inflection point? Try to answer this question (a) by graphing, (b) by using calculus.

Step-by-Step Solution

Answer

The graph of \(y = \sqrt{x} - \ln x\) has an inflection point at \(x = 4\).

Step 1: Understand the Function

The function given is \(y = \sqrt{x} - \ln x\) where \(x > 0\). This is a combination of the square root function and the natural logarithm with subtraction. We want to determine if the graph of this function has an inflection point.

Step 2: Graph the Function

Plot the function \(y = \sqrt{x} - \ln x\) for \(x > 0\). An inflection point is where the concavity of the curve changes, which you may identify by observing a change in the direction of curvature. Check if there are points where the curve transitions from concave up to concave down or vice versa.

Step 3: First Derivative

Find the first derivative of the function to understand its slope behavior. Given \(y = \sqrt{x} - \ln x\), we find \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{x}\). This derivative helps determine the intervals of increasing and decreasing behaviors, but doesn't directly show inflection points.

Step 4: Second Derivative

To find inflection points, compute the second derivative \(\frac{d^2y}{dx^2}\). Calculate \(\frac{d^2y}{dx^2} = -\frac{1}{4x^{3/2}} + \frac{1}{x^2}\). Set this second derivative equal to zero \(-\frac{1}{4x^{3/2}} + \frac{1}{x^2} = 0\) and solve for \(x\).

Step 5: Solve for Inflection Point

Solve \(-\frac{1}{4x^{3/2}} + \frac{1}{x^2} = 0\). Find a common denominator and simplify to get \(\frac{x^{3/2} - 4}{4x^{7/2}} = 0\). From this, solve \(x^{3/2} - 4 = 0\), yielding \(x = 8^{2/3} = 4\).

Step 6: Confirm Inflection Point

Check the sign of the second derivative around \(x = 4\) to ensure a change in concavity. For \(x < 4\), if \(\frac{d^2y}{dx^2}\) is positive and for \(x > 4\), if \(\frac{d^2y}{dx^2}\) is negative (or vice versa), there is an inflection point at \(x = 4\).

Key Concepts

Second Derivative

The second derivative is a powerful tool in calculus used to determine whether a function has an inflection point. It gives information about the concavity of a function's graph. When you calculate the second derivative of a function, you are assessing how the slope of the tangent line (given by the first derivative) changes.

For the function \(y = \sqrt{x} - \ln x\), the first derivative \(\frac{dy}{dx}\) is \(\frac{1}{2\sqrt{x}} - \frac{1}{x}\). This tells us about the function's rate of change. But to understand how the curve itself behaves, we compute the second derivative.

• The second derivative for our function is \(\frac{d^2y}{dx^2} = -\frac{1}{4x^{3/2}} + \frac{1}{x^2}\).
• Setting \(\frac{d^2y}{dx^2} = 0\) helps us find potential inflection points, as it indicates where the concavity might change.
• To solve, you find roots of \(-\frac{1}{4x^{3/2}} + \frac{1}{x^2} = 0\), simplifying to identify \(x = 4\).

The inflection point represents where the curve switches from bending one way to the other, which is a key visual in understanding a graph's behavior.

Concavity

Concavity describes the direction the curve of a function is facing—either up or down. A graph is concave up when it resembles a cup—like a smile—sometimes termed as convex, and concave down when it looks like an arch—like a frown.

To determine concavity, we use the second derivative.

• If \(\frac{d^2y}{dx^2} > 0\), the function is concave up at that interval.
• If \(\frac{d^2y}{dx^2} < 0\), it is concave down.

In our example function, \(y = \sqrt{x} - \ln x\), the computed second derivative changes sign around \(x = 4\). Before this point, if \(\frac{d^2y}{dx^2}\) is positive, the function is concave up, indicating the curve is shaped upwards. After \(x = 4\), if \(\frac{d^2y}{dx^2}\) becomes negative, the concavity is downward. This change in concavity confirms an inflection point at \(x = 4\), where the nature of the curve transitions.

Graphing Techniques

Graphing provides a visual means to understand and analyze the behavior of a function. With techniques of calculus, such as deriving and calculating derivatives, graphing becomes much clearer.

For the function \(y = \sqrt{x} - \ln x\):

• Begin by plotting basic points to get a general sense of the graph's shape.
• Use the first derivative to determine intervals where the function increases or decreases.
• The second derivative further assists in pinpointing where the graph curves, switching from concave up to concave down, as seen by the change in concavity at the inflection point \(x = 4\).

By identifying these elements on a graph, you can visually confirm the calculus-based findings. This dual approach—analytical and graphical—ensures robust understanding and verification of where and why curves change direction, highlighting key points such as inflection points.