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

Posted on May 31, 2020 by Myhometuition

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)

\begin{array}{l} y = 3 x (4 - x) \\ y = 12 x - 3 x^{2} \\ \frac{d y}{d x} = 12 - 6 x \\ When y is maximum, \frac{d y}{d x} = 0 \\ 0 = 12 - 6 x \\ x = 2 \end{array}

(b)

\begin{array}{l} y = 12 x - 3 x^{2} \\ When x = 2, \\ y = 12 (2) - 3 {(2)}^{2} \\ y = 12 \end{array}

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:

\begin{array}{l} y = 2 x^{3} + 3 x^{2} - 12 x + 7 \\ \frac{d y}{d x} = 6 x^{2} + 6 x - 12 \\ At turning point, \frac{d y}{d x} = 0 \end{array}

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.

\begin{array}{l} \frac{d^{2} y}{d x^{2}} = 12 x + 6 \\ When x = 1, \\ \frac{d^{2} y}{d x^{2}} = 12 (1) + 6 = 18 > 0 (positive) \end{array}

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

\begin{array}{l} When x = - 2, \\ \frac{d^{2} y}{d x^{2}} = 12 (- 2) + 6 = - 18 < 0 (negative) \end{array}

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

9.2.1 First Derivative for Polynomial Function (Examples)

Posted on May 24, 2020 by Myhometuition

Example:

Find

\frac{d y}{d x}

for each of the following functions:

(a) y = 12

(b) y = x⁴

(c) y = 3x

(d) y = 5x³

\begin{array}{l} (e) y = \frac{1}{x} \\ (f) y = \frac{2}{x^{4}} \\ (g) y = \frac{2}{5 x^{2}} \\ (h) y = 3 \sqrt{x} \\ (i) y = 4 \sqrt{x^{3}} \end{array}

Solution:

(a)
y = 12

\frac{d y}{d x} = 0

(b)
y = x⁴

\frac{d y}{d x}

= 4x³

(c)
y = 3x

\frac{d y}{d x}

= 3

(d)
y = 5x³

\frac{d y}{d x}

= 3(5x²) = 15x²

(e)

\begin{array}{l} y = \frac{1}{x} = x^{- 1} \\ \frac{d y}{d x} = - x^{- 1 - 1} = - \frac{1}{x^{2}} \end{array}

(f)

\begin{array}{l} y = \frac{2}{x^{4}} = 2 x^{- 4} \\ \frac{d y}{d x} = - 4 (2 x^{- 4 - 1}) = - 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{d y}{d x} = - 2 (\frac{2 x^{- 2 - 1}}{5}) = - \frac{4 x^{- 3}}{5} = - \frac{4}{5 x^{3}} \end{array}

(h)

\begin{array}{l} y = 3 \sqrt{x} = 3 {(x)}^{\frac{1}{2}} \\ \frac{d y}{d x} = \frac{1}{2} (3 x^{\frac{1}{2} - 1}) = \frac{3}{2} x^{- \frac{1}{2}} = \frac{3}{2 \sqrt{x}} \end{array}

(i)

\begin{array}{l} y = 4 \sqrt{x^{3}} = 4 {(x^{3})}^{\frac{1}{2}} = 4 x^{\frac{3}{2}} \\ \frac{d y}{d x} = \frac{3}{2} (4 x^{\frac{3}{2} - 1}) = 6 x^{\frac{1}{2}} = 6 \sqrt{x} \end{array}

Short Questions (Question 26 & 27)

Posted on May 18, 2020 by Myhometuition

Question 26 (3 marks):
Find the value of

\begin{array}{l} (a) \lim_{x \to 1} (7 - x^{2}), \\ (b) f'' (2) if f' (x) = 2 x^{3} - 4 x + 3. \end{array}

Solution:
(a)

\begin{array}{l} \lim_{x \to 1} (7 - x^{2}) \\ = 7 - {(1)}^{2} \\ = 6 \end{array}

(b)

\begin{array}{l} f' (x) = 2 x^{3} - 4 x + 3 \\ f'' (x) = 6 x^{2} - 4 \\ f'' (2) = 6 {(2)}^{2} - 4 \\ = 24 - 4 \\ = 20 \end{array}

Question 27 (4 marks):
It is given that L = 4t – t² and x = 3 + 6t.
(a) Express

\frac{d L}{d x}

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)

\begin{array}{l} Given L = 4 t - t^{2} and x = 3 + 6 t \\ L = 4 t - t^{2} \\ \frac{d L}{d t} = 4 - 2 t \\ x = 3 + 6 t \\ \frac{d x}{d t} = 6 \\ \frac{d L}{d x} = \frac{d L}{d t} \times \frac{d t}{d x} \\ \frac{d L}{d x} = (4 - 2 t) \times \frac{1}{6} \\ = \frac{4 - 2 t}{6} \\ = \frac{2 - t}{3} \end{array}

(b)

\begin{array}{l} δ L = 3.4 - 3 = 0.4 \\ \frac{δ L}{δ x} \approx \frac{d L}{d x} \\ δ x = δ L \div \frac{δ L}{δ x} \\ δ x = δ L \times \frac{δ x}{δ L} \\ = 0.4 \times \frac{3}{2 - t} \\ = \frac{2}{5} \times \frac{3}{2 - t} \\ = \frac{6}{5 (2 - t)} \\ When t = 1, \\ δ x = \frac{6}{5 (2 - 1)} = \frac{6}{5} \end{array}

Long Questions (Question 5)

Posted on May 17, 2020 by Myhometuition

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{d y}{d x}

when x = 3.
(b) Hence, estimate the value of

\frac{5}{{(2.98)}^{2}} .

Solution:
(a)

\begin{array}{l} y = \frac{5}{x^{2}} = 5 x^{- 2} \\ \frac{d y}{d x} = - 10 x^{- 3} \\ = - \frac{10}{x^{3}} \\ When x = 3 \\ \frac{d y}{d x} = - \frac{10}{3^{3}} = - \frac{10}{27} \end{array}

(b)

\begin{array}{l} δ x = 2.98 - 3 = - 0.02 \\ δ y = \frac{d y}{d x} . δ x \\ = - \frac{10}{27} \times (- 0.02) \\ = 0.007407 \\ Values of \frac{5}{{(2.98)}^{2}} \\ = y + δ y \\ = \frac{5}{x^{2}} + (0.007407) \\ = \frac{5}{3^{2}} + (0.007407) \\ = 0.56296 \end{array}

Long Questions (Question 4)

Posted on May 17, 2020 by Myhometuition

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:

\begin{array}{l} y = \frac{1}{64} x^{3} - \frac{3}{16} x^{2} ............... (1) \\ \frac{d y}{d x} = 3 (\frac{1}{64}) x^{2} - 2 (\frac{3}{16}) x^{1} \\ = \frac{3}{64} x^{2} - \frac{3}{8} x \\ At turning point, \frac{d y}{d x} = 0 \\ \frac{3}{64} x^{2} - \frac{3}{8} x = 0 \\ x (\frac{3}{64} x - \frac{3}{8}) = 0 \\ x = 0 or \\ \frac{3}{64} x - \frac{3}{8} = 0 \\ \frac{3}{64} x = \frac{3}{8} \\ x = \frac{3}{8} \times \frac{64}{3} \\ x = 8 \\ Substitute values of x into equation (1): \\ When x = 0, \\ y = \frac{1}{64} {(0)}^{3} - \frac{3}{16} {(0)}^{2} \\ y = 0 \\ When x = 8, \\ y = \frac{1}{64} {(8)}^{3} - \frac{3}{16} {(8)}^{2} \\ y = - 4 \\ Thus, turning points : (0, 0) and (8, - 4) \end{array}

\begin{array}{l} \frac{d y}{d x} = \frac{3}{64} x^{2} - \frac{3}{8} x \\ \frac{d^{2} y}{d x^{2}} = 2 (\frac{3}{64}) x - \frac{3}{8} \\ = \frac{3}{32} x - \frac{3}{8} \\ When x = 0, \\ \frac{d^{2} y}{d x^{2}} = \frac{3}{32} (0) - \frac{3}{8} \\ = - \frac{3}{8} (< 0) \\ (0, 0) is maximum point . \\ When x = 8, \\ \frac{d^{2} y}{d x^{2}} = \frac{3}{32} (8) - \frac{3}{8} \\ = \frac{3}{8} (> 0) \\ (8, - 4) is minimum point . \\ Shortest vertical distance between track \\ and ground level is at the minimum point . \\ Shortest vertical distance \\ = 5 - 4 \\ = 1 m \end{array}

Short Questions (Question 17 – 19)

Posted on May 16, 2020 by Myhometuition

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:

\begin{array}{l} y = \frac{x^{2}}{2} + 1 \\ \frac{d y}{d x} = x \\ At point Q, x -coordinate = 2, \\ Gradient of the curve, \frac{d y}{d x} = 2 \\ Hence, gradient of normal to the curve, P R = - \frac{1}{2} \\ \frac{3 - 0}{2 - k} = - \frac{1}{2} \\ 6 = - 2 + k \\ k = 8 \end{array}

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

$\frac{d y}{d x}$ = 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:

\begin{array}{l} y = \frac{3}{4} x^{2} \\ \frac{d y}{d x} = (2) \frac{3}{4} x = \frac{3}{2} x \\ δ y = 47.7 - 48 = - 0.3 \\ Approximate change in x to y \\ \frac{δ x}{δ y} \approx \frac{d x}{d y} \\ δ x = \frac{d x}{d y} \times δ y \\ δ x = \frac{2}{3 x} \times (- 0.3) \\ δ x = \frac{2}{3 (8)} \times (- 0.3) \leftarrow \begin{array}{l} y = 48 \\ \frac{3}{4} x^{2} = 48 \\ x^{2} = 64 \\ x = 8 \end{array} \\ δ x = - 0.025 \end{array}

Short Questions (Question 8 – 10)

Posted on May 16, 2020 by Myhometuition

Question 8:

Find

\frac{d s}{d t}

for each of the following functions.

\begin{array}{l} (a) s = {(t - \frac{3}{t})}^{2} \\ (b) s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \end{array}

Solution:
(a)

\begin{array}{l} s = {(t - \frac{3}{t})}^{2} \\ s = (t - \frac{3}{t}) (t - \frac{3}{t}) \\ s = t^{2} - 6 + \frac{9}{t^{2}} \\ s = t^{2} - 6 + 9 t^{- 2} \\ \frac{d s}{d t} = 2 t - 18 t^{- 3} = 2 t - \frac{18}{t^{3}} \end{array}

(b)

\begin{array}{l} s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \\ s = \frac{3 t - 5 t^{2} + 3 - 5 t}{t^{2}} = \frac{- 5 t^{2} - 2 t + 3}{t^{2}} \\ s = - 5 - \frac{2}{t} + \frac{3}{t^{2}} = - 5 - 2 t^{- 1} + 3 t^{- 2} \\ \frac{d s}{d t} = 2 t^{- 2} - 6 t^{- 3} = \frac{2}{t^{2}} - \frac{6}{t^{3}} \end{array}

Question 9:

Given that y = \frac{1 - 5 x^{4}}{x - 3}, find \frac{d y}{d x} .

Solution:

\begin{array}{l} \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(x - 3) . - 20 x^{3} - (1 - 5 x^{4}) .1}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 20 x^{4} + 60 x^{3} - 1 + 5 x^{4}}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 15 x^{4} + 60 x^{3} - 1}{{(x - 3)}^{2}} \end{array}

Question 10:

Given that f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x}, find f' (0) .

Solution:

\begin{array}{l} f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x} \\ f' (x) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(1 - 3 x) .5 {(x^{2} - 3)}^{4} .2 x - {(x^{2} - 3)}^{5} . - 3}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{10 x (1 - 3 x) {(x^{2} - 3)}^{4} + 3 {(x^{2} - 3)}^{5}}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [10 x - 30 x^{2} + 3 (x^{2} - 3)]}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [- 27 x^{2} + 10 x - 9]}{{(1 - 3 x)}^{2}} \\ ∴ f' (0) = \frac{{(0^{2} - 3)}^{4} [- 27 {(0)}^{2} + 10 (0) - 9]}{{(1 - 3 (0))}^{2}} \\ f' (0) = \frac{81 \times (- 9)}{1} = - 729 \end{array}

Long Questions (Question 2 & 3)

Posted on April 22, 2020 by user

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)

\begin{array}{l} y = x^{2} (x - 3) + 1 \\ y = x^{3} - 3 x^{2} + 1 \\ \frac{d y}{d x} = 3 x^{2} - 6 x \\ When x = - 1 \\ \frac{d y}{d x} = 3 {(- 1)}^{2} - 6 (- 1) \\ = 9 \\ Gradient of the curve is 9 . \end{array}

(b)
At turning points,

\frac{d y}{d x} = 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)

\begin{array}{l} y = 2 x {(1 - x)}^{4} \\ \frac{d y}{d x} = 2 x \times 4 {(1 - x)}^{3} (- 1) + {(1 - x)}^{4} \times 2 \\ = - 8 x {(1 - x)}^{3} + 2 {(1 - x)}^{4} \\ At T (2, 4), x = 2. \\ \frac{d y}{d x} = - 16 (- 1) + 2 (1) \\ = 16 + 2 \\ = 18 \end{array}

(b)

\begin{array}{l} Equation of normal: \\ y - y_{1} = - \frac{1}{\frac{d y}{d x}} (x - x_{1}) \\ y - 4 = - \frac{1}{18} (x - 2) \\ 18 y - 72 = - x + 2 \\ x + 18 y = 74 \end{array}

Long Questions (Question 1)

Posted on April 22, 2020 by user

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 B 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):

y – y₁= 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 B = (1, 7)

\begin{array}{l} when x = 1, \\ \frac{d^{2} y}{d x^{2}} = 6 x - 12 \\ \frac{d^{2} y}{d x^{2}} = 6 (1) - 12 \\ \frac{d^{2} y}{d x^{2}} = - 6 < 0 \\ Since \frac{d^{2} y}{d x^{2}} < 0, B is a maximum point . \end{array}

Short Questions (Question 22 – 25)

Posted on April 22, 2020 by user

Question 23:

Given that

y = 15 x + \frac{24}{x^{3}}

(a) Find the value of

\frac{d y}{d x}

when x = 2,

(b) Express in terms of k, the approximate change in y when x changes from 2 to

2 + k, where k is a small change.

Solution:
(a)

\begin{array}{l} y = 15 x + \frac{24}{x^{3}} \\ y = 15 x + 24 x^{- 3} \\ \frac{d y}{d x} = 15 - 72 x^{- 4} \\ \frac{d y}{d x} = 15 - \frac{72}{x^{4}} \\ When x = 2 \\ \frac{d y}{d x} = 15 - \frac{72}{2^{4}} = 10.5 \end{array}

(b)

\begin{array}{l} Approximate change in y to x in terms of k, \\ \frac{δ y}{δ x} \approx \frac{d y}{d x} \\ δ y = \frac{d y}{d x} \times δ x \\ δ y = 10.5 \times (2 + k - 2) \\ δ y = 10.5 k \end{array}

Question 24:

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

Solution:

\begin{array}{l} Area of circle, A = π r^{2} \\ \frac{d A}{d r} = 2 π r \\ Approximate increase in the area to radius, \\ \frac{δ A}{δ r} \approx \frac{d A}{d r} \\ δ A = \frac{d A}{d r} \times δ r \\ δ A = (2 π r) \times (4.01 - 4) \\ δ A = [2 π (4)] \times (0.01) \\ δ A = 0.08 π {cm}^{2} \end{array}

Question 25:

Given that y =3t+ 5t² and x = 5t –1.

(a) Find

\frac{d y}{d x}

in terms of x,

(b) If x increases from 5 to 5.01, find the small increase in t.

Solution:

\begin{array}{l} y = 3 t + 5 t^{2} \\ \frac{d y}{d t} = 3 + 10 t \\ x = 5 t - 1 \\ \frac{d x}{d t} = 5 \end{array}

(a)

\begin{array}{l} \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} = (3 + 10 t) \times \frac{1}{5} \\ \frac{d y}{d x} = \frac{3 + 10 (\frac{x + 1}{5})}{5} \leftarrow \begin{array}{l} x = 5 t - 1 \\ t = \frac{x + 1}{5} \end{array} \\ \frac{d y}{d x} = \frac{3 + 2 x + 2}{5} \\ \frac{d y}{d x} = \frac{5 + 2 x}{5} \end{array}

(b)

\begin{array}{l} Small increase in t to x, \\ δ t = \frac{d t}{d x} \times δ x \\ δ t = \frac{1}{5} \times (5.01 - 5) \\ δ t = 0.002 \end{array}