Calculus - Concavity – points of inflection

Given the function $f(x) = x^3 - 6x^2 + 9x + 5$, find the coordinates of the point of inflection and determine the interval where the function is concave up.

1. Find the first derivative: $f'(x) = 3x^2 - 12x + 9$

2. Find the second derivative: $f''(x) = 6x - 12$

3. Solve $f''(x) = 0$ for potential points of inflection: $6x - 12 = 0$ $6x = 12$ $x = 2$

4. Test the concavity around $x = 2$: For $x < 2$ (e.g., $x = 0$): $f''(0) = -12 < 0$ (Concave Down) For $x > 2$ (e.g., $x = 3$): $f''(3) = 6(3) - 12 = 6 > 0$ (Concave Up) Since the concavity changes at $x = 2$, it is a point of inflection.

5. Find the $y$-coordinate: $f(2) = (2)^3 - 6(2)^2 + 9(2) + 5$ $f(2) = 8 - 24 + 18 + 5 = 7$

6. The point of inflection is $(2, 7)$. The function is concave up for $x \in (2, \infty)$.

To find concavity, we calculate the second derivative. A point of inflection occurs where $f''(x)$ changes sign. By setting $f''(x) = 0$, we find the critical value $x=2$ and use a sign test to confirm the change from concave down to concave up.

Show that the function $f(x) = x^4$ has $f''(0) = 0$ but does not have a point of inflection at $x = 0$.

1. First derivative: $f'(x) = 4x^3$
2. Second derivative: $f''(x) = 12x^2$
3. Set second derivative to zero: $12x^2 = 0 \Rightarrow x = 0$.
4. Test the sign of $f''(x)$ around $x = 0$:
   • For $x < 0$ (e.g., $x = -1$): $f''(-1) = 12(-1)^2 = 12 > 0$ (Concave Up)
   • For $x > 0$ (e.g., $x = 1$): $f''(1) = 12(1)^2 = 12 > 0$ (Concave Up)
5. Because $f''(x)$ does not change sign (it is positive on both sides of $x = 0$), there is no point of inflection at $x = 0$.

This example illustrates that $f''(x) = 0$ is a necessary but not sufficient condition for a point of inflection. A change in the sign of the second derivative must also occur.