# 9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points (Examples)

Example 1 (Maximum Value of Quadratic Function)
Given that y = 3x (4 – x), calculate
(a) the value of x when y is a maximum,
(b) the maximum value of y.

Solution:
(a)

(b)

Example 2 (Determine the Turning Points and Second Derivative Test)
Find the coordinates of the turning points on the curve y = 2x3 + 3x2 – 12x + 7 and determine the nature of these turning points.

Solution:

6x2 + 6x – 12 = 0
x2 + x – 2 = 0
(x – 1) (x + 2) = 0
x = 1 or x = –2

When x = 1
y = 2(1)3 + 3(1)2 – 12(11) + 7
y = 0
(1, 0) is a turning point.

When x = –2
y = 2(–2)3 + 3(–2)2 – 12(–2) + 7
y = 27
(–2, 27) is a turning point.

Hence, the turning point (1, 0) is a minimum point.

Hence, the turning point (–2, 27) is a maximum point.

# 9.2.1 First Derivative for Polynomial Function (Examples)

Example:
Find $\frac{dy}{dx}$ for each of the following functions:
(a) y = 12
(b) y = x4
(c) y = 3x
(d) y = 5x3

Solution:
(a)
= 12
$\frac{dy}{dx}=0$

(b)
= x4
$\frac{dy}{dx}$ = 4x3

(c)
= 3x
$\frac{dy}{dx}$ = 3

(d)
= 5x3
$\frac{dy}{dx}$ = 3(5x2) = 15x2

(e)

$\begin{array}{l}y=\frac{1}{x}={x}^{-1}\\ \frac{dy}{dx}=-{x}^{-1-1}=-\frac{1}{{x}^{2}}\end{array}$

(f)

$\begin{array}{l}y=\frac{2}{{x}^{4}}=2{x}^{-4}\\ \frac{dy}{dx}=-4\left(2{x}^{-4-1}\right)=-8{x}^{-5}=-\frac{8}{{x}^{5}}\end{array}$

(g)

$\begin{array}{l}y=\frac{2}{5{x}^{2}}=\frac{2{x}^{-2}}{5}\\ \frac{dy}{dx}=-2\left(\frac{2{x}^{-2-1}}{5}\right)=-\frac{4{x}^{-3}}{5}=-\frac{4}{5{x}^{3}}\end{array}$

(h)

$\begin{array}{l}y=\text{3}\sqrt{x}=3{\left(x\right)}^{\frac{1}{2}}\\ \frac{dy}{dx}=\frac{1}{2}\left(3{x}^{\frac{1}{2}-1}\right)=\frac{3}{2}{x}^{-\frac{1}{2}}=\frac{3}{2\sqrt{x}}\end{array}$

(i)
$\begin{array}{l}y=4\sqrt{{x}^{3}}=4{\left({x}^{3}\right)}^{\frac{1}{2}}=4{x}^{\frac{3}{2}}\\ \frac{dy}{dx}=\frac{3}{2}\left(4{x}^{\frac{3}{2}-1}\right)=6{x}^{\frac{1}{2}}=6\sqrt{x}\end{array}$

# Short Questions (Question 26 & 27)

Question 26 (3 marks):
Find the value of

Solution:
(a)
$\begin{array}{l}\underset{x\to 1}{\mathrm{lim}}\left(7-{x}^{2}\right)\\ =7-{\left(1\right)}^{2}\\ =6\end{array}$

(b)

Question 27 (4 marks):
It is given that L = 4t t2 and x = 3 + 6t.
(a) Express $\frac{dL}{dx}$ in terms of t.
(b) Find the small change in x, when L changes from 3 to 3.4 at the instant t = 1.

Solution:
(a)

(b)

# Long Questions (Question 5)

Question 5 (7 marks):
It is given that the equation of a curve is $y=\frac{5}{{x}^{2}}.$
(a) Find the value of $\frac{dy}{dx}$ when x = 3.
(b) Hence, estimate the value of $\frac{5}{{\left(2.98\right)}^{2}}.$

Solution:
(a)

(b)

# Long Questions (Question 4)

Question 4 (6 marks):
Diagram shows the front view of a part of a roller coaster track in a miniature park.

The curve part of the track of the roller coaster is represented by an equation $y=\frac{1}{64}{x}^{3}-\frac{3}{16}{x}^{2}$ , with point A as the region.
Find the shortest vertical distance, in m, from the track to ground level.

Solution:

# Short Questions (Question 17 – 19)

Question 17:

In the diagram above, the straight line PR is normal to the curve   $y=\frac{{x}^{2}}{2}+1$ at Q. Find the value of k.

Solution:

Question 18:
The normal to the curve y = x2 + 3x at the point P is parallel to the straight line
y = x + 12. Find the equation of the normal to the curve at the point P.

Solution:
Given normal to the curve at point P is parallel to the straight line y = –x + 12.
Hence, gradient of normal to the curve = –1.
As a result, gradient of tangent to the curve = 1

y = x2 + 3x
$\frac{dy}{dx}$ = 2x + 3
2x + 3 = 1
2x = –2
x = –1
y = (–1)2+ 3(–1)
y = –2
Point P = (–1, –2).

Equation of the normal to the curve at point P is,
y – (–2) = –1 (x – (–1))
y + 2 = – x – 1
y = – x– 3

Question 19:
Given that   $y=\frac{3}{4}{x}^{2}$ , find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:

# Short Questions (Question 8 – 10)

Question 8:
Find $\frac{ds}{dt}$ for each of the following functions.

Solution:
(a)
$\begin{array}{l}s={\left(t-\frac{3}{t}\right)}^{2}\\ s=\left(t-\frac{3}{t}\right)\left(t-\frac{3}{t}\right)\\ s={t}^{2}-6+\frac{9}{{t}^{2}}\\ s={t}^{2}-6+9{t}^{-2}\\ \frac{ds}{dt}=2t-18{t}^{-3}=2t-\frac{18}{{t}^{3}}\end{array}$

(b)
$\begin{array}{l}s=\frac{\left(t+1\right)\left(3-5t\right)}{{t}^{2}}\\ s=\frac{3t-5{t}^{2}+3-5t}{{t}^{2}}=\frac{-5{t}^{2}-2t+3}{{t}^{2}}\\ s=-5-\frac{2}{t}+\frac{3}{{t}^{2}}=-5-2{t}^{-1}+3{t}^{-2}\\ \frac{ds}{dt}=2{t}^{-2}-6{t}^{-3}=\frac{2}{{t}^{2}}-\frac{6}{{t}^{3}}\end{array}$

Question 9:

Solution:
$\begin{array}{l}\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{v}^{2}}=\frac{\left(x-3\right).-20{x}^{3}-\left(1-5{x}^{4}\right).1}{{\left(x-3\right)}^{2}}\\ \frac{dy}{dx}=\frac{-20{x}^{4}+60{x}^{3}-1+5{x}^{4}}{{\left(x-3\right)}^{2}}\\ \frac{dy}{dx}=\frac{-15{x}^{4}+60{x}^{3}-1}{{\left(x-3\right)}^{2}}\end{array}$

Question 10:

Solution:
$\begin{array}{l}f\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{5}}{1-3x}\\ f\text{'}\left(x\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{v}^{2}}=\frac{\left(1-3x\right).5{\left({x}^{2}-3\right)}^{4}.2x-{\left({x}^{2}-3\right)}^{5}.-3}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{10x\left(1-3x\right){\left({x}^{2}-3\right)}^{4}+3{\left({x}^{2}-3\right)}^{5}}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{4}\left[10x-30{x}^{2}+3\left({x}^{2}-3\right)\right]}{{\left(1-3x\right)}^{2}}\\ f\text{'}\left(x\right)=\frac{{\left({x}^{2}-3\right)}^{4}\left[-27{x}^{2}+10x-9\right]}{{\left(1-3x\right)}^{2}}\\ \\ \therefore f\text{'}\left(0\right)=\frac{{\left({0}^{2}-3\right)}^{4}\left[-27{\left(0\right)}^{2}+10\left(0\right)-9\right]}{{\left(1-3\left(0\right)\right)}^{2}}\\ f\text{'}\left(0\right)=\frac{81×\left(-9\right)}{1}=-729\end{array}$

# Long Questions (Question 2 & 3)

Question 2:
Given the equation of a curve is:
y = x2 (x – 3) + 1
(a) Find the gradient of the curve at the point where x = –1.
(b) Find the coordinates of the turning points.

Solution:
(a)

(b)
At turning points, $\frac{dy}{dx}=0$
3x2 – 6x = 0
x2 – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x2 (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 22 (2 – 3) + 1
y = 4 (–1) + 1 = –3
Therefore, coordinates of the turning points are (0, 1) and (2, –3).

Question 3:
It is given the equation of the curve is y = 2x (1 – x)4 and the curve pass through T(2, 4).
Find
(a) the gradient of the curve at point T.
(b) the equation of the normal to the curve at point T.

Solution:
(a)

(b)

# Long Questions (Question 1)

Question 1:
The curve y = x3 – 6x2 + 9x + 3 passes through the point P (2, 5) and has two turning points, A (3, 3) and B.
Find
(a) the gradient of the curve at P.
(b) the equation of the normal to the curve at P.
(c) the coordinates of and determine whether B is a maximum or the minimum point.

Solution:
(a)
y = x3 – 6x2 + 9x + 3
dy/dx = 3x2– 12x + 9
At point P (2, 5),
dy/dx = 3(2)2 – 12(2) + 9 = –3

Gradient of the curve at point P = –3.

(b)
Gradient of normal at point P = 1/3
Equation of the normal at P (2, 5):
yy1 = m (x – x1)
y – 5 = 1/3 (x – 2)
3y – 15 = x – 2
3y = x + 13

(c)
At turning point, dy/dx = 0.
3x2 – 12x + 9 = 0
x2 – 4x + 3 = 0
(x – 1)( x – 3) = 0
x – 1 = 0  or x – 3 = 0
x = 1  x = 3 (Point A)

Thus at point B:
x = 1
y = (1)3– 6(1)2 + 9(1) + 3 = 7

Thus, coordinates of = (1, 7)

# Short Questions (Question 22 – 25)

Question 23:
Given that $y=15x+\frac{24}{{x}^{3}}$ ,

(a) Find the value of $\frac{dy}{dx}$ when x = 2,
(b) Express in terms of k, the approximate change in when x changes from 2 to
2 + k, where k is a small change.

Solution:
(a)

(b)

Question 24:
If the radius of a circle increases from 4 cm to 4.01 cm, find the approximate increase in the area.

Solution:

Question 25:
Given that y =3t+ 5t2 and x = 5t 1.
(a) Find $\frac{dy}{dx}$  in terms of x,
(b) If increases from 5 to 5.01, find the small increase in t.

Solution:
$\begin{array}{l}y=3t+5{t}^{2}\\ \frac{dy}{dt}=3+10t\\ \\ x=5t-1\\ \frac{dx}{dt}=5\end{array}$

(a)
$\begin{array}{l}\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}\\ \frac{dy}{dx}=\left(3+10t\right)×\frac{1}{5}\\ \frac{dy}{dx}=\frac{3+10\left(\frac{x+1}{5}\right)}{5}←\overline{)\begin{array}{l}x=5t-1\\ t=\frac{x+1}{5}\end{array}}\\ \frac{dy}{dx}=\frac{3+2x+2}{5}\\ \frac{dy}{dx}=\frac{5+2x}{5}\end{array}$

(b)