Problem 11 - Essential Calculus

Question

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=x(x-4)^{3}\)

Step-by-Step Solution

Answer

The points of inflection of the function are at \(x=0\), \(x=4\), and \(x=-6\). The graph of the function is concave down on \((-∞, -6)\) and \((0, 4)\), and is concave up on \((-6, 0)\) and \((4, ∞)\).

Step 1: Find the derivative

By the product rule and the chain rule, the derivative is \( f'(x)=3x(x-4)^{2}+x(x-4)^{3}=x(x-4)^{2}(3+x) \)

Step 2: Identify critical points

Critical points occur where \(f'(x)=0\) or \(f'(x)\) is undefined. Here, the derivative is defined for all x, so we solve \(f'(x)=0\) for x, which gives us \(x=0\), \(x=4\), and \(x=-3\).

Step 3: Find the second derivative

Now by applying the product rule and chain rule again, the second derivative is \(f''(x)=2x(x-4)(3+x)+3(x-4)^{2}(3+x)+3x(x-4)^{2}=x(x-4)(6+x)\)

Step 4: Identify Points of inflection

Points of inflection occur where \(f''(x)=0\) or \(f''(x)\) is undefined. Here, the second derivative \(f''(x)\) is defined for all x, so we solve the equation \(f''(x)=0\) to get the values \(x=0\), \(x=4\) and \(x=-6\).

Step 5: Discuss the concavity

The sign of \(f''(x)\) tells us whether \(f(x)\) is concave up (\(f''(x)>0\)) or concave down (\(f''(x)<0\)). So we test values in each interval formed by the points of inflection, for example, for \(x<-6\), \(f''(x)<0\); for \(-60\); for \(04\), \(f''(x)>0\). Therefore, the graph is concave down on \((-∞, -6)\) and \((0, 4)\), and is concave up on \((-6, 0)\) and \((4, ∞)\).