Problem 14 - Essential Calculus

Question

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=\frac{x+1}{\sqrt{x}}\)

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x+1}{\sqrt{x}}\) has an inflection point at \(x= \frac{2}{3}\). For \(0 < x < \frac{2}{3}\), the function is concave down, and for \(x > \frac{2}{3}\), the function is concave up.

Step 1: Find the first derivative

Using the quotient rule for derivation, get the derivative of \(f(x)\):\n\(f'(x)= \frac{\sqrt{x}(1) - (x+1)(\frac{1}{2\sqrt{x}})}{[\sqrt{x}]^2}\) that simplifies to:\n \(f'(x)=\frac{1}{2\sqrt{x}} - \frac{ x+1}{2x}\)

Step 2: Simplify the first derivative

Simplify the first derivative to have common denominators. The result is \(f'(x)=\frac{-x+2}{2x\sqrt{x}}\)

Step 3: Find the second derivative

Differentiate \(f'(x)\) to obtain the second derivative. Using the quotient rule again, we get\n\(f''(x)= \frac{2x\sqrt{x} - (-x+2)\frac{1}{2\sqrt{x}}}{4x^2} \)\n that simplifies to:\n\( f''(x) = \frac{3x-2}{4x^2\sqrt{x}} \)

Step 4: Find points of inflection

Setting \( f''(x) = 0 \) will give us potential inflection points.\nThe equation \(3x-2 = 0\) leads to \(x = \frac{2}{3}\). The second derivative is also undefined at \(x = 0\), however \(x = 0\) is not in the domain of the original function and thus cannot be a point of inflection.

Step 5: Discuss the concavity

Now make a sign chart for \( f''(x) \). Divide the domain of \( f \) into intervals using the potential inflection point \(x = \frac{2}{3}\).For \(0 < x < \frac{2}{3}\), \( f''(x) \) is negative, so \( f \) is concave down. For \(x > \frac{2}{3}\), \( f''(x) \) is positive, so \( f \) is concave up. Therefore, \( x = \frac{2}{3} \) is indeed a point of inflection as the concavity changes here.