1.7.5 Locating and Determining the Nature of Stationary Points

In this topic we will learn how to:

locate stationary points and determine their nature using the second derivative

$\textbf{\large\textcolor{gray}{Stationary Points}}$ A stationary point is a point on the graph with a gradient of $0$ . This means that,
$\frac{dy}{dx} = 0$ To determine the nature of a stationary point i.e whether its a maximum point or a minimum point, we have to find the second derivative. If the second derivative is negative, then it is a maximum point. If the second derivative is positive, it is a minimum point.

Let’s look at some past paper questions.

1. A curve has equation $f$ which is defined by $f(x) = \frac{2}{3}x^{3} - 7x + \frac{4}{x} + 2$ , and it is given that $f'(x) = 2x^{2} - 7 - \frac{4}{x^{2}}$ .

(a) Find the coordinates of the stationary point on the curve.
$f'(x) = 2x^{2} - 7 - \frac{4}{x^{2}}$ $f'(x)$ represents the derivative of $f(x)$ i.e it is the same as $\frac{dy}{dx}$ . Therefore, since we already have the gradient function let’s equate it to $0$ ,
$2x^{2} - 7 - \frac{4}{x^{2}} = 0$ Multiply through by $x^{2}$ ,
$2x^{4} - 7x^{2} - 4 = 0$ Recognise and solve the hidden quadratic,
$2x^{4} - 7x^{2} - 4 = 0$ $2(x^{2})^{2} - 7x^{2} - 4 = 0$ $\textbf{Let } y = x^{2}$ $2y^{2} - 7y - 4 = 0$ $(2y + 1)(y - 4) = 0$ $2y + 1 = 0\ \ \ \ \ \ \ \ y - 4 = 0$ $y = -\frac{1}{2}\ \ \ \ \ \ \ \ y = 4$ $y = x^{2}$ $x^{2} = -\frac{1}{2}\ \ \ \ \ \ \ \ x^{2} = 4$ $\sqrt{x^{2}} = \sqrt{-\frac{1}{2}}\ \ \ \ \ \ \ \ \sqrt{x^{2}} = \pm\sqrt{4}$ $x = \textmd{no solutions}\ \ \ \ \ \ \ \ x = \pm 2$ Note: The square root of a negative number has no real roots, hence no solutions for $x = \sqrt{-\frac{1}{2}}$ .

Now that we have the $x$ -coordinates, let’s find the $y$ -coordinates by substituting into $f(x)$ ,
$f(x) = \frac{2}{3}x^{3} - 7x + \frac{4}{x} + 2$ $\textmd{at } x = -2\ \ \ \ \ \ \ \ \textmd{at } x = 2$ $f(-2) = \frac{2}{3}(-2)^{3} - 7(-2) + \frac{4}{(-2)} + 2\ \ \ \ \ \ \ \ f(2) = \frac{2}{3}(2)^{3} - 7(2) + \frac{4}{(2)} + 2$ $f(-2) = \frac{26}{3}\ \ \ \ \ \ \ \ f(2) = -\frac{14}{3}$ Therefore, the coordinates of the stationary points are,
$\left(-2, \frac{26}{3}\right) \left(2, -\frac{14}{3}\right)$ (b) Find $f''(x)$ .
$f'(x) = 2x^{2} - 7 - \frac{4}{x^{2}}$ Differentiate $f'(x)$ ,
$f'(x) = 2x^{2} - 7 - 4x^{-2}$ $f''(x) = 4x - (4)(-2)x^{-3}$ $f''(x) = 4x + 8x^{-3}$ Therefore, the final answer is,
$f''(x) = 4x + 8x^{-3}$ (c) Hence, or otherwise, determine the nature of the stationary points.

To determine the nature of the stationary points, we use the second derivative,
$f''(x) = 4x + 8x^{-3}$ Substitute the $x$ -coordinates at the stationary points,
$\textmd{at } \left(-2, \frac{26}{3}\right)\ \ \ \ \ \ \ \ \textmd{at } \left(2, -\frac{14}{3}\right)$ $f''(-2) = 4(-2) + 8(-2)^{-3}\ \ \ \ \ \ \ \ f''(2) = 4(2) + 8(2)^{-3}$ $f''(-2) = -9\ \ \ \ \ \ \ \ f''(2) = 9$ From the information above we can tell that,
$f''(-2) < 0 \textmd{ hence maximum point}$ $f''(2) > 0 \textmd{ hence minimum point}$ Therefore, the final answer is,
$\textmd{At }\left(-2, \frac{26}{3}\right)\textmd{, }f''(x) < 0\textmd{, hence it is a maximum point.}$ $\textmd{At }\left(2, -\frac{14}{3}\right)\textmd{, }f''(x) > 0\textmd{, hence it is a minimum point.}$ 2. The equation of a curve is $y = (3 - 2x)^{3} + 24x$ .

(a) Find the expressions for $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ .

Let’s start by finding $\frac{dy}{dx}$ ,
$y = (3 - 2x)^{3} + 24x$ $\frac{dy}{dx} = (3)(-2)(3 - 2x)^{2} + 24$ $\frac{dy}{dx} = -6(3 - 2x)^{2} + 24$ Now let’s find $\frac{d^{2}y}{dx^{2}}$ ,
$\frac{dy}{dx} = -6(3 - 2x)^{2} + 24$ $\frac{d^{2}y}{dx^{2}} = (-6)(2)(3 - 2x)$ $\frac{d^{2}y}{dx^{2}} = -12(3 - 2x)$ $\frac{d^{2}y}{dx^{2}} = -36 + 24x$ Therefore, the final answer is,
$\frac{dy}{dx} = -6(3 - 2x)^{2} + 24\ \ \ \ \ \ \ \ \frac{d^{2}y}{dx^{2}} = -36 + 24x$ (b) Find the coordinates of each of the stationary point on the curve.

$\frac{dy}{dx} = -6(3 - 2x)^{2} + 24$ At a stationary point $\frac{dy}{dx} = 0$ ,
$-6(3 - 2x)^{2} + 24 = 0$ Solve for $x$ ,
$6(3 - 2x)^{2} = 24$ Divide both sides by $6$ ,
$(3 - 2x)^{2} = \frac{24}{6}$ $(3 - 2x)^{2} = 4$ Take the square root of both sides,
$\sqrt{(3 - 2x)^{2}} = \pm \sqrt{4}$ $3 - 2x = \pm 2$ Make $x$ the subject of the formula,
$3 - 2x = \pm 2$ $2x = 3 \pm 2$ $x = \frac{3}{2} \pm \frac{2}{2}$ $x = \frac{3}{2} \pm 1$ $x = \frac{3}{2} - 1\ \ \ \ \ \ \ \ x = \frac{3}{2} + 1$ $x = \frac{1}{2}\ \ \ \ \ \ \ \ x = \frac{5}{2}$ Substitute the $x$ -coordinates into the original equation to find the $y$ -coordinates,
$y = (3 - 2x)^{3} + 24x$ $\textmd{at } x = \frac{1}{2}\ \ \ \ \ \ \ \ \textmd{at } x = \frac{5}{2}$ $y = \left(3 - 2\left(\frac{1}{2}\right)\right)^{3} + 24\left(\frac{1}{2}\right)\ \ \ \ \ \ \ \ y = \left(3 - 2\left(\frac{5}{2}\right)\right)^{3} + 24\left(\frac{5}{2}\right)$ $y = 20\ \ \ \ \ \ \ \ y = 52$ Therefore, the coordinates of the stationary points are,
$\left(\frac{1}{2}, 20\right) \left(\frac{5}{2}, 52\right)$ (c) Determine the nature of the stationary points.

To determine the nature of the stationary points, we use the second derivative,
$\frac{d^{2}y}{dx^{2}} = -36 + 24x$ Substitute the $x$ -coordinates of the stationary points into the second derivative,
$\frac{d^{2}y}{dx^{2}} = -36 + 24x$ $\textmd{at } \left(\frac{1}{2}, 20\right)\ \ \ \ \ \ \ \ \textmd{at } \left(\frac{5}{2}, 52\right)$ $\frac{d^{2}y}{dx^{2}} = -36 + 24\left(\frac{1}{2}\right)\ \ \ \ \ \ \ \ \frac{d^{2}y}{dx^{2}} = -36 + 24\left(\frac{5}{2}\right)$ $\frac{d^{2}y}{dx^{2}} = -24\ \ \ \ \ \ \ \ \frac{d^{2}y}{dx^{2}} = 24$ $\frac{d^{2}y}{dx^{2}} < 0 \textmd{ hence maximum point}\ \ \ \ \ \ \ \ \frac{d^{2}y}{dx^{2}} > 0 \textmd{ hence minimum point}$ Therefore, the final answer is,
$\textmd{At }\left(\frac{1}{2}, 20\right)\textmd{, }\frac{d^{2}y}{dx^{2}} < 0\textmd{, hence it is a maximum point.}$ $\textmd{At }\left(\frac{1}{2}, 20\right)\textmd{, }\frac{d^{2}y}{dx^{2}} > 0\textmd{, hence it is a minimum point.}$

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