How to Find Point of Inflection?

To find a point of inflection, you systematically analyze the second derivative of a function to determine where its concavity changes.

Understanding Points of Inflection

A point of inflection on the graph of a function is a point where the function's concavity changes. This means the curve switches from being concave up (like a cup opening upwards) to concave down (like a cup opening downwards), or vice versa. For a point to be an inflection point, the function must be continuous at that point.

Step-by-Step Guide to Finding Inflection Points

The most common and reliable method to find points of inflection involves using the second derivative of the function. This method helps identify where the rate of change of the slope itself changes direction.

Step 1: Compute the Second Derivative

Begin by finding the first derivative of your function, denoted as $f'(x)$. Then, differentiate $f'(x)$ to find the second derivative, denoted as $f''(x)$. The second derivative provides information about the concavity of the original function.

Step 2: Identify Potential Inflection Points

Once you have $f''(x)$, the next crucial step is to find the values of $x$ where:

• $f''(x) = 0$
• $f''(x)$ does not exist (e.g., due to division by zero or a discontinuity).

These values are called possible inflection points or critical points for the second derivative. These are the only places where the concavity could change.

Step 3: Test for Concavity Change

The final step is to determine if the concavity actually changes at these potential points. This is done by creating test intervals using the $x$-values found in Step 2 and then evaluating the sign of $f''(x)$ within each interval.

1. Create Test Intervals: Use the $x$-values identified in Step 2 to divide the number line into intervals.
2. Choose Test Values: Select a test value within each interval.
3. Evaluate $f''(x)$: Substitute each test value into $f''(x)$ and observe the sign:
   • If $f''(x) > 0$ on an interval, the function is concave up on that interval.
   • If $f''(x) < 0$ on an interval, the function is concave down on that interval.
4. Confirm Inflection Points: An actual inflection point occurs at an $x$-value if:
   • $f''(x)$ changes sign (from positive to negative or negative to positive) when crossing that $x$-value.
   • The original function $f(x)$ is defined at that $x$-value.

If the sign of $f''(x)$ does not change across a potential point, then it is not an inflection point, even if $f''(x)=0$ or is undefined there.

Example: Finding an Inflection Point

Let's find the inflection point(s) for the function $f(x) = x^3 - 3x^2 + 2$.

1. Find the first derivative:
   $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2) = 3x^2 - 6x$

2. Find the second derivative:
   $f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$

3. Find potential inflection points:
   Set $f''(x) = 0$:
   $6x - 6 = 0$
   $6x = 6$
   $x = 1$
   Since $f''(x)$ is a polynomial, it exists everywhere, so $x=1$ is our only potential inflection point.

4. Test intervals for concavity change:
   We use $x=1$ to create two intervals: $(-\infty, 1)$ and $(1, \infty)$.

   Interval	Test Value ($x$)	$f''(x) = 6x - 6$	Sign of $f''(x)$	Concavity
   $(-\infty, 1)$	$x=0$	$6(0) - 6 = -6$	Negative (-)	Concave Down
   $(1, \infty)$	$x=2$	$6(2) - 6 = 6$	Positive (+)	Concave Up

   Since the sign of $f''(x)$ changes from negative to positive at $x=1$, there is an inflection point at $x=1$.

5. Find the y-coordinate of the inflection point:
   Substitute $x=1$ back into the original function $f(x)$:
   $f(1) = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$

Therefore, the point of inflection is (1, 0).