Gradients and Stationary Points

Gradients of Tangents to Curves

The derivative of a function at a particular value of \(x\) gives the gradient of the tangent to the graph at that point.

Example

A curve has equation

\( y = x - \frac{16}{\sqrt{x}} \)

Find the gradient of the tangent at \(x = 4\).

\[ y = x - \frac{16}{\sqrt{x}} \] \[ y = x - 16x^{-1/2} \] \[ \frac{dy}{dx} = 1 - \left(-\frac12 \cdot 16 x^{-3/2}\right) \] \[ \frac{dy}{dx} = 1 + \frac{8}{\sqrt{x^3}} \]

\[ \text{when } x = 4: \quad \frac{dy}{dx} = 1 + \frac{8}{\sqrt{4^3}} \] \[ = 1 + \frac{8}{\sqrt{64}} \] \[ = 1 + \frac{8}{8} \] \[ = 2 \]

The gradient of the equation when x = 4 is 2

Finding the Equation of a Tangent

Example

The point \(A(-1, 7)\) lies on the curve: \( y = 5x^2 + 2 \)

Find the equation of the tangent at point A.

\[ y = 5x^2 + 2 \] \[ \frac{dy}{dx} = 10x \] \[ \text{when } x = -1: \quad \frac{dy}{dx} = -10 \] \[ \text{Tangent has gradient } -10 \]

\[ \text{Point } A(-1, 7) \text{ lies on the curve} \] \[ y - b = m(x - a) \] \[ y - 7 = -10\bigl(x - (-1)\bigr) \] \[ y - 7 = -10(x + 1) \] \[ y - 7 = -10x - 10 \] \[ y = -10x - 3 \]

The equation of the tangent to the curve \( y = 5x^2 + 2 \) at point A is \( y = -10x - 3 \)

Increasing and Decreasing Functions

The graph below is decreasing when \(x \lt 0\):

The value of the function decreases as \(x\) increases.
The gradient is negative.

It is increasing when \(x > 0\):

The value of the function increases as \(x\) increases.
The gradient is positive.

When \(x = 0\), the gradient is zero and the graph changes from negative to positive gradient.

This turning point is called a stationary point.

The stationary point can be:

Maximum
Minimum
Rising point of inflection
Falling point of inflection

Finding Stationary Points

Stationary points occur when \(\frac{dy}{dx} = 0\)

or \(f'(x) = 0\)

Example

Find the stationary points of the equation \[ y = x^3 - 3x^2 +8 \]

\[ y = x^3 - 3x^2 + 8 \] \[ \frac{dy}{dx} = 3x^2 - 6x \] \[ \text{Stationary points occur when } \frac{dy}{dx} = 0 \] \[ 0 = 3x^2 - 6x \] \[ 0 = 3x(x - 2) \] \[ x = 0 \quad \text{or} \quad x = 2 \]

\[ y = x^3 - 3x^2 + 8 \] \[ \text{when } x = 0: \quad y = 0^3 - 3(0^2) + 8 = 8 \] \[ \text{when } x = 2: \quad y = 2^3 - 3(2^2) + 8 \] \[ y = 8 - 12 + 8 = 4 \] \[ \text{Stationary points are } (0, 8) \text{ and } (2, 4) \]

Nature Tables

A nature table shows how a function behaves on either side of its stationary points.

A stationary point occurs when \(f'(x)=0\). To classify it, examine the sign of \(f'(x)\) just before and after the point.

\( f'(x) \) changes from positive to negative → maximum
\( f'(x) \) changes from negative to positive → minimum
\( f'(x) \) does not change sign → stationary point of inflection

To find out if the stationary point is a maximum, minimum or point of inflection, construct a nature table:-

Put in the values of x for the stationary points.
Copy these values, with a small minus and plus sign.
Copy the first part of the factorised form of the derivative.
Repeat for subsequent parts of the factorised form of the derivative.
Write down the whole factorised form of the derivative.

Example

Find the nature of the turning points of the equation \[ y = x^3 - 3x^2 +8 \]

From the work above \[ y = x^3 - 3x^2 + 8 \] \[ \frac{dy}{dx} = 3x^2 - 6x \] \[ \text{Set } \frac{dy}{dx} = 0: \quad 3x^2 - 6x = 0 \] \[ 3x(x - 2) = 0 \] \[ x = 0 \quad \text{or} \quad x = 2 \] \[ \text{When } x = 0: \quad y = 0^3 - 3(0^2) + 8 = 8 \] \[ \text{When } x = 2: \quad y = 2^3 - 3(2^2) + 8 = 4 \] \[ \text{Stationary points: } (0, 8) \text{ and } (2, 4) \]

Examine what happens near \( x = 0\) and \( x = 2 \)

Remember that \( - \times - \) is a +

So if \( 3x \) is negative and \((x - 2)\) ia also negative, \( 3x(x-2) \) is positive.

Example

Find the nature of the turning points of the equation \( f(x) = x^3 - 3x^2 - 4x + 12 \) .

\[ f(x) = x^3 - 3x^2 - 4x + 12 \] \[ f'(x) = 3x^2 - 6x - 4 \] \[ \text{Set } f'(x) = 0: \quad 3x^2 - 6x - 4 = 0 \] \[ 3x^2 - 6x - 4 = (3x + 2)(x - 2) \] \[ (3x + 2)(x - 2) = 0 \] \[ 3x + 2 = 0 \quad \text{or} \quad x - 2 = 0 \] \[ x = -\frac{2}{3} \quad \text{or} \quad x = 2 \] \[ \text{When } x = -\frac{2}{3}: \quad f\!\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 - 3\left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) + 12 = \frac{352}{27} \] \[ \text{When } x = 2: \quad f(2) = 2^3 - 3(2^2) - 4(2) + 12 = 0 \] \[ \text{Turning points: } \left(-\frac{2}{3}, \frac{352}{27}\right) \text{ and } (2, 0) \]

\[ f'(x) = 3x^2 - 6x - 4 = (3x+2)(x-2) \] \[ \begin{array}{c|c|c|c} x & x \lt -\frac{2}{3} & -\frac{2}{3} \lt x \lt 2 & x > 2 \\ \hline 3x+2 & - & + & + \\ x-2 & - & - & + \\ \hline f'(x) & + & - & + \\ \end{array} \] \[ \begin{array}{c|c|c} \text{Interval} & \text{Sign of } f'(x) & \text{Nature of } f(x) \\ \hline x \lt -\frac{2}{3} & + & \text{Increasing} \\ -\frac{2}{3} \lt x \lt 2 & - & \text{Decreasing} \\ x > 2 & + & \text{Increasing} \\ \end{array} \] \[ \text{So } x = -\frac{2}{3} \text{ is a local maximum, and } x = 2 \text{ is a local minimum.} \] \[ \left(-\frac{2}{3}, \frac{352}{27}\right) \text{ is a local maximum, } \text{ and } (2, 0) \text{ is a local minimum.} \]

Second Derivative

The second derivative \(f''(x)\) measures how the gradient changes with respect to \(x\).

When \(f''(x) > 0\), the curve is concave up → minimum.

When \(f''(x) \lt 0\), the curve is concave down → maximum.

When \(f''(x) = 0\), there may be a point of inflection — confirm with a nature table.

Example

\[ f(x)=x^3 - 3x^2 + 8 \] \[ f'(x)=3x^2 - 6x \] \[ f''(x)=6x - 6 \] \[ \text{Stationary points when } f'(x)=0 \] \[ 0 = 3x^2 - 6x \] \[ 0 = 3x(x-2) \] \[ x = 0 \qquad \text{or} \qquad x = 2 \] \[ f(0)=8 \qquad\qquad f(2)=8 - 12 + 8 = 4 \] \[ f''(0) = -6 \qquad\qquad f''(2)=6 \] \[ (0,8)\ \text{is a maximum} \qquad\qquad (2,4)\ \text{is a minimum} \]

Closed Intervals

In a closed interval, the maximum and minimum values of a function occur either at a stationary point or at an endpoint.

Example

\[ f(x)=x^3 - 3x^2 + 8,\qquad -1.5 \le x \le 2.5 \]

Stationary points occur when \(f'(x)=0\).

\[ f(x)=x^3 - 3x^2 + 8 \] \[ \Rightarrow\ f'(x)=3x^2 - 6x \] \[ \text{st. pts: } 0 = 3x^2 - 6x \] \[ \Rightarrow\ 0 = x(3x - 6) \] \[ \Rightarrow\ x = 0,\; 2 \]

\[ \text{When } x = 0 \] \[ f(0)=0^3 - 3\cdot 0^2 + 8 = 8 \] \[ f''(x)=6x - 6 \] \[ f''(0) = -6 \] \[ \text{Maximum turning point at } (0,8) \] \[ \text{since } f''(0) \lt 0 \]

\[ \text{When } x = 2 \] \[ f(2)=2^3 - 3\cdot 2^2 + 8 = 4 \] \[ f''(x)=6x - 6 \] \[ f''(2)=6 \] \[ \text{Minimum turning point at } (2,4) \] \[ \text{since } f''(2) > 0 \]

Now check endpoints:

\[ f(x)=x^3 - 3x^2 + 8 \] \[ f(-1.5) = (-1.5)^3 - 3(-1.5)^2 + 8 \] \[ = -3.375 - 6.75 + 8 \] \[ = -2.125 \]
\[ f(2.5) = (2.5)^3 - 3(2.5)^2 + 8 \] \[ = 15.625 - 18.75 + 8 \] \[ = 4.875 \]

Maximum value: \(8\)
Minimum value: \(-2.125\)

Critical Points

Assuming \(a\) is in the domain of \(f\):

\[ \text{Critical points of a function occur when} \] \[ f'(a)=0 \] \[ \text{or when } f'(a) \text{ does not exist.} \]

\[ \text{If } (a,\,f(a)) \text{ is a critical point,} \] \[ a \text{ is called a critical number of the function.} \] \[ f(a) \text{ is called a critical value of the function.} \]

\[ \text{A local maximum value } f(a) \text{ exists at } a \text{ if, and only if,} \] \[ \text{there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \ge f(x) \text{ for all } x \text{ in the interval.} \]

\[ \text{A local minimum value } f(a) \text{ exists at } a \text{ if, and only if,} \] \[ \text{there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \le f(x) \text{ for all } x \text{ in the interval.} \]

Endpoints (also critical points):

\[ \text{An endpoint maximum value } f(a) \text{ exists at } a \] \[ \text{if } (a,\,f(a)) \text{ is an endpoint} \] \[ \text{and there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \ge f(x) \text{ for all } x \text{ in the interval.} \]

\[ \text{An endpoint minimum value } f(a) \text{ exists at } a \] \[ \text{if } (a,\,f(a)) \text{ is an endpoint} \] \[ \text{and there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \le f(x) \text{ for all } x \text{ in the interval.} \]

Global maxima and minima:

\[ \text{A global maximum value } f(a) \text{ exists at } a \] \[ \text{if, and only if, there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \ge f(x) \text{ for all } x \text{ in the domain.} \] \[ \text{i.e. it is the largest value for the domain being considered.} \]

\[ \text{A global minimum value } f(a) \text{ exists at } a \] \[ \text{if, and only if, there is an interval centred at } a \] \[ \text{in the domain such that } f(a) \le f(x) \text{ for all } x \text{ in the domain.} \] \[ \text{i.e. it is the smallest value for the domain being considered.} \]

A continuous function on a closed interval must have both a global maximum and minimum.
They are found by examining the local extrema and end points

Find the critical points.
List the endpoints.
The global extrema must be one of these.

Returning to the earlier example:

\[ f(x)=x^3 - 3x^2 + 8,\qquad -1.5 \le x \le 2.5 \]

\[ \text{Global max at } x=0,\qquad \text{Global min at } x=-1.5 \]

Example

Find the global extrema of the following piecewise function on the domain \([-3,2]\):

\[ f(x)= \begin{cases} x+5 & \text{when } -3 \le x \lt -2 \\[6pt] x^{2}-1 & \text{when } -2 \le x \lt -1 \\[6pt] 1 - x^{2} & \text{when } -1 \le x \lt 1 \\[6pt] x^{2}-1 & \text{when } 1 \le x \lt 2 \end{cases} \]

So:

\[ f(x)= \begin{cases} x+5 & \text{when } -3 \le x \lt -2 \\[6pt] x^{2}-1 & \text{when } -2 \le x \lt -1 \\[6pt] 1 - x^{2} & \text{when } -1 \le x \lt 1 \\[6pt] x^{2}-1 & \text{when } 1 \le x \lt 2 \end{cases} \]

\[ \text{Critical points occur when } f'(a)=0 \] \[ \text{or when } f'(a) \text{ does not exist for the point } (a,\,f(a)). \] \[ x = 0 \text{ is a stationary point.} \] \[ f'(x) = -2x, \qquad f''(0) = -2, \] \[ \text{so } (0,1) \text{ is a local maximum.} \]

\[ \text{The critical points to consider are } -3,\,-2,\,-1,\,1,\text{ and }2. \] \[ x=-3 \text{ is an endpoint with no left derivative.} \] \[ f(-3)=2,\qquad f(-2)=3 \] \[ (-3,2) \text{ is an endpoint minimum since it is an endpoint and has the smallest value on } -3 \le x \lt 2. \] \[ x=-2 \text{ has a left derivative of } 1 \text{ and a right derivative of } -4. \] \[ f(-2)=3,\qquad f(-1)=2 \] \[ (-2,3) \text{ is a local maximum.} \] \[ x=-1 \text{ has a left derivative of } -2 \text{ and a right derivative of } 2. \] \[ f(-1)=0,\qquad f(0)=1 \] \[ (-1,0) \text{ is a local minimum.} \] \[ x=1 \text{ has a left derivative of } -2 \text{ and a right derivative of } 2. \] \[ f(1)=0,\qquad f(1.99)=2.9601 \] \[ (1,0) \text{ is a local minimum.} \] \[ x=2 \text{ has no right derivative.} \] \[ \text{It is not an endpoint since it is excluded from the interval.} \]