Points of Inflection (Edexcel A-Level Mathematics): Revision Notes

1. Find the second derivative of the function, $f''(x)$.
2. Solve $f''(x) = 0$ to find potential inflection points.
3. Check for a change in concavity by determining the sign of $f''(x)$ before and after each potential inflection point.

• If $f''(x)$ changes sign around $x = c, then\ x = c$ is an inflection point.
• If $f''(x)$ does not change sign, then $x = c$ is not an inflection point.

Example 1: Find the Points of Inflection for $f(x) = x^3 - 3x^2 + 4x$

• Step 1: Find the first derivative: $f'(x) = 3x^2 - 6x + 4$

• Step 2: Find the second derivative: $f''(x) = 6x - 6$

• Step 3: Solve $f''(x) = 0$ to find potential inflection points: $6x - 6 = 0 \quad \Rightarrow \quad x = 1$

• Step 4: Check for a change in concavity:
• For $x < 1 (e.g., x = 0): f''(0) = 6(0) - 6 = -6$ (concave down).
• For $x > 1 (e.g., x = 2): f''(2) = 6(2) - 6 = 6$ (concave up).

• Conclusion: Since $f''(x)$ changes sign from negative to positive at $x = 1$, there is a point of inflection at $x = 1.$

Example 2: Find the Points of Inflection for $f(x) = x^4 - 4x^3 + 6x^2$

• Step 1: Find the first derivative: $f'(x) = 4x^3 - 12x^2 + 12x$

• Step 2: Find the second derivative: $f''(x) = 12x^2 - 24x + 12$

• Step 3: Solve f''(x) = 0 to find potential inflection points: $12x^2 - 24x + 12 = 0$ $x^2 - 2x + 1 = 0$ $(x - 1)^2 = 0$ $x = 1$

• Step 4: Check for a change in concavity:
• At $x = 1$,$f''(x)$ is $0$ but does not change sign (as$f''(x) = 12(x - 1)^2,$ which is always non-negative).

• Conclusion: Since $f''(x)$ does not change sign at $x = 1$, there is no point of inflection.

Example Visualization:

Consider the function $f(x) = x^3$. Its first derivative is $f'(x) = 3x^2,$ and its second derivative is $f''(x) = 6x.$

• $f''(x) > 0 for x > 0$ (concave up).
• $f''(x) < 0 for x < 0$ (concave down).
• The point $x$ $= 0$ is a point of inflection, where the graph changes concavity.

• Points of inflection are points where the graph of a function changes concavity.
• To find them, use the second derivative $f''(x)$, find where it equals zero or is undefined, and check if the sign changes around those points.
• Points of inflection are crucial in understanding the overall shape and behaviour of functions, with wide applications in mathematics, physics, economics, and engineering.

Example:

Example:

Drawing lines from the curve upwards, we see that:

• From $0 \leq x \leq \pi$, the curve is concave.
• From $\pi \leq x \leq 2\pi$, the curve is convex.

• A curve is concave in an interval if, for all values in that interval, $\frac{d^2y}{dx^2} < 0$. concave implies underground (i.e., < $$0)

• A curve is convex if $\frac{d^2y}{dx^2} > 0$ in an interval.

At this point, the curve stops being concave, and after this point, the curve starts being convex, or vice versa where the curve turns. Fact:

• Point of inflection $\Rightarrow f''(x) = 0$
• Point of inflection $\nLeftarrow f''(x) = 0$

Examples:

1. $y = x^4$ [Graph of $y = x^4$ showing no point of inflection]

2. $y = x^3$ [Graph of $y = x^3$ showing a point of inflection at the origin]

Finding the points where $\frac{d^2y}{dx^2} = 0$

Example

For each of the following functions, find the interval on which the function is:

4. Convex
5. Concave a. $f'(x) = x^3 - 3x^2 + x - 2$

• $f'(x) = 3x^2 - 6x + 1$
• $f''(x) = 6x - 6$

6. For convex, $6x - 6 > 0 \Rightarrow 6x > 6 \Rightarrow x > 1$

• Convex when x > 1

7. For concave, $6x - 6 < 0 \Rightarrow x < 1$

• Concave for x < 1