Fundamentals of Calculus Revision Notes for Leaving Cert Mathematics

Maximum and Minimum Turning Points

What are turning points?

Turning points are special locations on a curve where the function changes from increasing to decreasing, or vice versa. At these points, the tangent to the curve is horizontal, meaning it has zero slope.

At points A and B shown above, the tangent lines are parallel to the x-axis, which means their slope equals zero. Point A represents a maximum turning point (or local maximum), while point B represents a minimum turning point (or local minimum).

chatImportant

The key mathematical fact to remember:

At any turning point on a curve, $\frac{dy}{dx} = 0$

Finding turning points step-by-step

To locate the turning points of any function, follow this systematic approach:

Step 1: Find the first derivative $\frac{dy}{dx}$

Step 2: Set $\frac{dy}{dx} = 0$ and solve this equation to find the x-values

Step 3: Substitute these x-values back into the original function to find the corresponding y-values

Step 4: If there are multiple turning points, the one with the higher y-value is the maximum, and the one with the lower y-value is the minimum

Worked example 1: Finding a minimum turning point

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Worked Example: Finding a Minimum Turning Point

Let's find the turning point of $f(x) = x^2 - 2x - 5$ .

Step 1: Find the derivative $f'(x) = 2x - 2$

Step 2: Set the derivative equal to zero $2x - 2 = 0$ $2x = 2$ $x = 1$

Step 3: Find the y-coordinate When $x = 1$ : $y = (1)^2 - 2(1) - 5 = 1 - 2 - 5 = -6$

Therefore, the minimum turning point is $(1, -6)$

Worked example 2: Multiple turning points

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Worked Example: Finding Multiple Turning Points

Find the coordinates of the local maximum and minimum of $y = x^3 - 9x^2 + 15x + 2$ .

Step 1: Find $\frac{dy}{dx}$ $y = x^3 - 9x^2 + 15x + 2$ $\frac{dy}{dx} = 3x^2 - 18x + 15$

Step 2: Set $\frac{dy}{dx} = 0$ $3x^2 - 18x + 15 = 0$ $x^2 - 6x + 5 = 0$ (dividing by 3) $(x - 1)(x - 5) = 0$ Therefore: $x = 1$ and $x = 5$

Step 3: Find y-coordinates When $x = 1$ : $y = (1)^3 - 9(1)^2 + 15(1) + 2 = 1 - 9 + 15 + 2 = 9$ When $x = 5$ : $y = (5)^3 - 9(5)^2 + 15(5) + 2 = 125 - 225 + 75 + 2 = -23$

Step 4: Identify which is maximum and minimum Since $(1, 9)$ is higher than $(5, -23)$ :

$(1, 9)$ is the maximum turning point
$(5, -23)$ is the minimum turning point

The second derivative test

The second derivative helps us determine whether a turning point is a maximum or minimum without needing to compare y-values.

For any function $y = f(x)$ :

The first derivative is $\frac{dy}{dx}$ or $f'(x)$
The second derivative is $\frac{d^2y}{dx^2}$ or $f''(x)$

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The Second Derivative Test:

At a maximum turning point: $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} < 0$ (negative)

At a minimum turning point: $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} > 0$ (positive)

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Memory Aid: Think of it this way: at a maximum, the curve is "frowning" (∩ shape), so the second derivative is negative. At a minimum, the curve is "smiling" (∪ shape), so the second derivative is positive.

Worked example 3: Using the second derivative test

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Worked Example: Using the Second Derivative Test

Find the turning points of $y = x^3 - 3x^2 + 5$ and determine their nature.

Step 1: Find the first derivative $\frac{dy}{dx} = 3x^2 - 6x$

Step 2: Find turning points $3x^2 - 6x = 0$ $3x(x - 2) = 0$ Therefore: $x = 0$ or $x = 2$

Step 3: Find y-coordinates When $x = 0$ : $y = (0)^3 - 3(0)^2 + 5 = 5$ → Point $(0, 5)$ When $x = 2$ : $y = (2)^3 - 3(2)^2 + 5 = 8 - 12 + 5 = 1$ → Point $(2, 1)$

Step 4: Find the second derivative $\frac{d^2y}{dx^2} = 6x - 6$

Step 5: Apply the second derivative test At $x = 0$ : $\frac{d^2y}{dx^2} = 6(0) - 6 = -6$ (negative) Since the second derivative is negative, $(0, 5)$ is a maximum turning point

At $x = 2$ : $\frac{d^2y}{dx^2} = 6(2) - 6 = 6$ (positive) Since the second derivative is positive, $(2, 1)$ is a minimum turning point

Special case: Quadratic functions

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Quadratic Functions Special Case

Quadratic functions of the form $y = ax^2 + bx + c$ have exactly one turning point.

Important rule:

If $a > 0$ : the parabola opens upward (∪), so it has a minimum turning point
If $a < 0$ : the parabola opens downward (∩), so it has a maximum turning point

Increasing and decreasing functions

Understanding when a function increases or decreases is closely related to turning points.

When a curve is increasing: $\frac{dy}{dx} > 0$ (positive slope)

When a curve is decreasing: $\frac{dy}{dx} < 0$ (negative slope)

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Worked Example: Finding Where a Function Increases

For what values of x is the curve $y = 2x^2 + 8x - 5$ increasing?

Step 1: Find the derivative $\frac{dy}{dx} = 4x + 8$

Step 2: For the curve to be increasing, we need $\frac{dy}{dx} > 0$ $4x + 8 > 0$ $4x > -8$ $x > -2$

Therefore, the curve is increasing when $x > -2$

Key formulas summary

Concept	Formula	Meaning
Turning point condition	$\frac{dy}{dx} = 0$	Slope is zero
Maximum test	$\frac{d^2y}{dx^2} < 0$	Second derivative negative
Minimum test	$\frac{d^2y}{dx^2} > 0$	Second derivative positive
Increasing function	$\frac{dy}{dx} > 0$	Positive slope
Decreasing function	$\frac{dy}{dx} < 0$	Negative slope

bookmarkSummary

Key Points to Remember:

Turning points occur where $\frac{dy}{dx} = 0$ - this is the fundamental condition
Use the second derivative test to determine if a point is maximum ( $\frac{d^2y}{dx^2} < 0$ ) or minimum ( $\frac{d^2y}{dx^2} > 0$ )
Quadratic functions ( $y = ax^2 + bx + c$ ) have one turning point: minimum if $a > 0$ , maximum if $a < 0$
The function increases when $\frac{dy}{dx} > 0$ and decreases when $\frac{dy}{dx} < 0$
Always substitute x-values back into the original function to find the complete coordinates of turning points

Fundamentals of Calculus (Leaving Cert Mathematics): Revision Notes

Maximum and Minimum Turning Points

What are turning points?

Finding turning points step-by-step

Worked example 1: Finding a minimum turning point

Worked example 2: Multiple turning points

The second derivative test

Worked example 3: Using the second derivative test

Special case: Quadratic functions

Increasing and decreasing functions

Key formulas summary

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