To find stationary points of a function $f(x)$, follow these steps:

1. Find the first derivative $f'(x)$.
2. Set the first derivative equal to zero and solve for $x$:

3. Find the corresponding y-coordinates by substituting the values of $x$ back into the original function $f(x)$.
4. Classify the stationary points using the second derivative test or first derivative test.

• Local Maximum: The graph changes from increasing to decreasing at a turning point.
• Local Minimum: The graph changes from decreasing to increasing at a turning point.

a) Second Derivative Test:

The second derivative test can help classify stationary points:

• If $f''(x) > 0$ at a stationary point $x = c$, then $f(x)$ has a local minimum at $x = c$.
• If $f''(x) < 0$ at a stationary point $x = c$, then $f(x)$ has a local maximum at $x = c$.
• If $f''(x) = 0$ at a stationary point $x = c$, the test is inconclusive, and you may need to use higher derivatives or another method to classify the point.

b) First Derivative Test:

The first derivative test involves checking the sign of $f'(x)$ before and after the stationary point:

• If $f'(x)$ changes from positive to negative as $x$ passes through the stationary point, then $f(x)$ has a local maximum at that point.
• If $f'(x)$ changes from negative to positive, then $f(x)$ has a local minimum.
• If $f'(x)$ does not change sign (e.g., it remains positive or negative), the stationary point is a saddle point (not a turning point).

Example 1: Find and classify the stationary points of $f(x) = x^3 - 3x^2 + 2$.

• Step 1: Find the first derivative:

• Step 2: Find the stationary points by setting $f'(x) = 0$:

So, $x = 0$ and $x = 2$ are stationary points.

• Step 3: Classify the stationary points using the second derivative test.
• First, find the second derivative:

• Evaluate $f''(x)$ at the stationary points:
• At $x = 0: f''(0) = 6(0) - 6 = -6,\ so\ x = 0$ is a local maximum.
• At $x = 2: f''(2) = 6(2) - 6 = 6,\ so\ x = 2$ is a local minimum.
• Step 4: Find the corresponding y-values:
• At $x = 0: f(0) = 0^3 - 3(0)^2 + 2 = 2$.
• At $x = 2: f(2) = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$.

• Conclusion:
• The stationary point at (0, 2) is a local maximum.
• The stationary point at (2, -2) is a local minimum.

Example 2: Find and classify the stationary points of $f(x) = x^4 - 4x^3 + 4x^2$.

• Step 1: Find the first derivative:

• Step 2: Find the stationary points by setting $f'(x) = 0$:

So, $x = 0, x = 1$, and $x = 2$ are stationary points.

• Step 3: Classify the stationary points using the second derivative test.
• First, find the second derivative:

• Evaluate $f''(x)$ at the stationary points:
• At $x = 0: f''(0) = 12(0)^2 - 24(0) + 8 = 8$ (local minimum).
• At $x = 1: f''(1) = 12(1)^2 - 24(1) + 8 = -4$ (local maximum).
• At $x = 2: f''(2) = 12(2)^2 - 24(2) + 8 = 8$ (local minimum).

• Step 4: Find the corresponding $y$-values:
• At $x = 0: f(0) = 0^4 - 4(0)^3 + 4(0)^2 = 0$.
• At $x = 1: f(1) = 1^4 - 4(1)^3 + 4(1)^2 = 1 - 4 + 4 = 1$.
• At $x = 2: f(2) = 2^4 - 4(2)^3 + 4(2)^2 = 16 - 32 + 16 = 0$.

• Conclusion:
• The stationary point at (0, 0) is a local minimum.
• The stationary point at (1, 1) is a local maximum.
• The stationary point at (2, 0) is a local minimum.

• Stationary points occur where the first derivative of a function is zero, and they can be classified as local maxima, minima, or saddle points.
• Turning points are specific types of stationary points where the function changes direction, marking local maxima or minima.
• The second derivative test helps in classifying these points by determining the concavity of the function around them.
• Understanding stationary and turning points is crucial for analysing and interpreting the behaviour of functions in various mathematical and real-world contexts.

$S_1$ is called a local maximum.

• "Maximum" because it is the highest point in the locality.
• "Local" because it is not the maximum of the entire graph. $S_2$ is called a local minimum for similar reasons.

A stationary point that is the max/min of the entire function is called a Global max/min.

Example: Find the coordinates of the stationary points of $y = x^3 - 12x$.

1. Differentiate to find the gradient function.

2. Solve for $\dfrac{dy}{dx} = 0$:

3. If asked, go back to coordinate equations to find y coordinates:

• When $x = 2$:

• When $x = -2$:

At a known stationary point:

• If $\dfrac{d^2y}{dx^2} > 0 \Rightarrow$ Min
• If $\dfrac{d^2y}{dx^2} < 0 \Rightarrow$ Max
• If $\dfrac{d^2y}{dx^2} = 0 \Rightarrow$ No information $\therefore$ have to experiment numerically.

Example: Find and classify the stationary points of $y = x^3 - 12x$.

1. Find $\dfrac{dy}{dx} = 3x^2 - 12 = 0 \Rightarrow$ (Worth 1 mark)

2. When $x = 2, y = (2)^3 - 12(2) = -16 \Rightarrow (2, -16)$. When $x = -2, y = (-2)^3 - 12(-2) = 16 \Rightarrow (-2, 16)$.

Now, classifying the points:

• $\dfrac{d^2y}{dx^2} = 6x$ At $x = 2, \dfrac{d^2y}{dx^2} = 6(2) = 12 > 0 \Rightarrow$ Min at (2, -16). (Worth a mark)

At $x = -2, \dfrac{d^2y}{dx^2} = 6(-2) = -12 < 0 \Rightarrow$ Max at (-2, 16).

Example: Find and classify any stationary points of $y = x^4 + 6$

1. $\dfrac{dy}{dx} = 4x^3 = 0 \Rightarrow x = 0, y = (0)^4 + 6 = 6 \Rightarrow (0, 6)$
2. $\dfrac{d^2y}{dx^2} = 12x^2,\ which\ at\ x = 0\ gives\ \dfrac{d^2y}{dx^2} = 12(0)^2 = 0$ When this happens, we must test the gradient slightly to the left and slightly to the right of the stationary point:

• Left: $f'(-0.1) = \dfrac{-1}{250}$
• Right: $f'(0.1) = \dfrac{1}{250}$
• $\text{ then } + \Rightarrow \text{ Min}$

Find and classify the stat. point of $y = x^3 + 2$

1. $\dfrac{dy}{dx} = 3x^2 = 0 \Rightarrow x = 0, y = 2 \Rightarrow (0, 2)$
2. $\dfrac{d^2y}{dx^2} = 6x$, which at $x = 0$ gives $\dfrac{d^2y}{dx^2} = 6(0) = 0$ Left: $f'(-0.1) = 3(-0.1)^2 = 0.03 > 0$

Right: $f'(0.1) = 3(0.1)^2 = 0.03 > 0$

• $\oplus \ to \ \oplus \Rightarrow \text{ point of inflection} \ ( \text{same for } \ominus \ to \ \ominus )$