Problem 12 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The points of inflection are \(x = 3\). The function is concave down on the interval \((-1, 3)\) and concave up on the interval \((3, \infty)\).

Step 1: Compute the first derivative

Use the product rule and the chain rule to differentiate. The product and chain rules state that \((g(x)h(x))' = g'(x)h(x) + g(x)h'(x)\) and \((f(g(x))' = f'(g(x))g'(x)\) respectively. So, you get \(f'(x) = \sqrt{x+1} + x(1/(2\sqrt{x+1}))\).

Step 2: Compute the second derivative

Again using the product, chain, and power rules, the second derivative of \(f(x)\) can be found as \(f''(x) = (1/(2\sqrt{x+1})) - (x/(4(x+1)^{(3/2)}))\).

Step 3: Find points of inflection

Points of inflection are values of \(x\) for which the second derivative equals zero or is undefined. Set \(f''(x)\) equal to zero and solve for \(x\), and also check where \(f''(x)\) is undefined. The solutions form the critical points: \(x = -1, 3\). Note that \(x = -1\) has to be discarded as it lies outside the domain of the original function \(f(x)\).

Step 4: Test the concavity

To determine which intervals are concave up or down, choose a test point in each interval determined by the points of inflection and substitute into the second derivative. If \(f''(x) > 0\), the interval is concave up. If \(f''(x) < 0\), the interval is concave down. Thus, the function is concave down on \((-1, 3)\) and concave up on \((3, \infty)\).