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,
$$\textmd{At }\left(\frac{1}{2}, 20\right)\textmd{, }\frac{d^{2}y}{dx^{2}} < 0\textmd{, hence it is a maximum point.}$$$$\textmd{At }\left(\frac{1}{2}, 20\right)\textmd{, }\frac{d^{2}y}{dx^{2}} > 0\textmd{, hence it is a minimum point.}$$