Let's consider the function \(f(x) = 2x^3 - 9x^2 + 12x - 3\). Find the intervals of concavity and the inflection points.
Answer
The inflection point is then at \(x = 1.5\), and we find the corresponding y-value by substituting \(x = 1.5\) into the original function: \(f(1.5) = 2(1.5)^3 - 9(1.5)^2 + 12(1.5) - 3 = -1.125\). Therefore, the inflection point is \((1.5, -1.125)\)
Step 1 ：First, we find the first derivative of the function: \(f'(x) = 6x^2 - 18x + 12\)
Step 2 ：Then, we find the second derivative of the function: \(f''(x) = 12x - 18\)
Step 3 ：Set the second derivative equal to zero and solve for x: \(12x - 18 = 0\) which gives \(x = 1.5\)
Step 4 ：To find the intervals of concavity, we test the sign of the second derivative on each interval. Choose test points \(x = 1\) and \(x = 2\). For \(x = 1\), \(f''(1) = -6 < 0\), so the function is concave down on the interval \((-\infty, 1.5)\). For \(x = 2\), \(f''(2) = 6 > 0\), so the function is concave up on the interval \((1.5, \infty)\)
Step 5 ：The inflection point is then at \(x = 1.5\), and we find the corresponding y-value by substituting \(x = 1.5\) into the original function: \(f(1.5) = 2(1.5)^3 - 9(1.5)^2 + 12(1.5) - 3 = -1.125\). Therefore, the inflection point is \((1.5, -1.125)\)
