MATH 1090 Algebra and Trigonometry
Problem Set 4 - Solutions

2002 Winter

Note

Without the WP MathA font, the symbol “” (set of real numbers) appears as “ú”.

The point P has coordinates (2, -1). Find:
1. the equation of the line of slope 4 that passes through P;
  Point-slope form of line equation:
  y - (-1) = 4 (x - 2)
  [which is acceptable as a final answer.]
  Converting to slope-intercept form (which has a much more appealing appearance):
  y + 1 = 4 x - 8
  Þ
  
  y = 4 x - 9
2. the equation of the line perpendicular to the line in part (a) and passing through the point (2, 3); and
  
  Therefore the equation of the line, in general form (and avoiding non-integer coefficients) is
  
  x + 4y - 14 = 0
3. the equation of the circle of radius 4 and centre at P.
  Standard form is (x - h)² + (y - k)² = r²
  (h, k) = (2, -1) and r = 4
  Þ
  
  (x - 2)² + (y + 1)² = 16
  
  Converting to general form:
  (x² - 4x + 4) + (y² + 2y + 1) - 16 = 0
  Þ x² + y² - 4x + 2y - 11 = 0
4. Sketch all three objects on the same diagram.

Sketch, on the same diagram, the curve and line whose equations are given by
1. x² + y² + 6x - 8y = 0
  Completing the square,
  (x + 3)² - 9 + (y - 4)² - 16 = 0
  Þ (x + 3)² + (y - 4)² = 9 + 16 = 25 = 5²
  Therefore the circle has its centre at (-3, 4) and has radius 5.
  Also note that x = y = 0 satisfies the equation of the circle, so that it must pass through the origin.
2. 4x - 3y + 24 = 0
  x = 0 Þ 0 + - 3y + 24 = 0 Þ 3y = 24
  Therefore y-intercept is 8.
  y = 0 Þ 4x + 0 + 24 = 0 Þ 4x = -24
  Therefore x-intercept is -6.
  This is sufficient information to sketch the line.
  
  Note that the line cuts the circle at both of the line’s axis intercepts.

For the function y = x³ - x ,

Sketch the curve from x = -2 to x = 2.

First note that y is an odd function of x.
[It is a polynomial with only odd exponents present - also see part (c) below.]
Therefore we need sketch only half of the curve.
The other half will follow by symmetry about the origin.

x	x³	y = x³ - x
0.0	0.000	0.000
0.5	0.125	-0.375
1.0	1.000	0.000
1.5	3.375	1.875
2.0	8.000	6.000

One can plot additional points if needed.
[Graph of y = x^3 - x]

Use the vertical line test to illustrate that f (x) is a function.
No matter where we place a vertical line on this graph, the line cuts the curve only once.

Therefore f (x) is a function.

Is f (x) even, odd or neither?
As noted in part (a) above,

f (x) is odd.

Proof:
f (-x) = (-x)³ - (-x) = -x³ + x = -(x³ - x) = -f (x) for all x.

Find the x and y intercepts.

x intercepts occur where y = 0 :
y = x³ - x = x (x² - 1) = x (x - 1) (x + 1)
y = 0 Þ x = 0 or 1 or -1 .

y intercepts occur where x = 0 :
f (0) = 0³ - 0 = 0
Therefore

The x-intercepts are -1, 0, 1
and the only y-intercept is 0 .

Find the domain and range of f (x).
f (x) is a polynomial. The domain can be extended beyond the interval requested in part (a) to include all real numbers. The leading term has an odd exponent, so the function has no upper or lower bounds (it “goes to both + and -”).
Therefore

The domain and range are both ú

Find the average rate of change of f (x) over each of the intervals
1. x = 1 to x = 2
2. x = 1 to x = 1.1
3. x = 1 to x = 1.01
The average rate of change from x₁ to x₂ is .
Therefore
1. From x = 1 to x = 2 :
2. From x = 1 to x = 1.1 :
3. From x = 1 to x = 1.01 :

Challenge questions:

Find the average rate of change of f (x) over the interval from x = 1 to x = 1 + h , (h ¹ 0). Hence find the limit of the sequence of average rates of change that you found in part (f), as the second point approaches the first point at x = 1 .
$[f(1+h) - f(1)] / [(1+h) - 1] = [{(1+h)^3 - (1+h)} - {1^3 - 1}] / h = (1 + 3h + 3h^2 + h^3 - 0) / h = h(2 + 3h + h^2)/h = 2 + 3h + h^2$
[Note: by substituting 1, 0.1 and 0.01 for h in each of parts (f)(i), (ii) and (iii) respectively, one can verify the answers for part (f).]
Therefore the average rate of change is

2 + 3h + h² , (h ¹ 0)

Clearly, as h ® 0, the average rate of change at x = 1 tends to

2

Generalize part (g) by finding and simplifying the difference quotient and find its limiting value as h ® 0.
[This is the instantaneous rate of change of f (x) at any point (x, f (x)) on the curve.]
$[f(x+h) - f(x)] / [(x+h) - x] = [{(x+h)^3 - (x+h)} - {x^3 - x}] / h = (x^3 + 3x^2 h + 3x h^2 + h^3 - x - h - x^3 + x) / h = h(3x^2 - 1 + 3xh + h^2)/h =$

3 x² - 1 + 3xh + h²

Clearly, as h ® 0, the average rate of change at x tends to

3 x² - 1

Find the domain and range of the function

In general the domain is the set of all real numbers except those causing division by zero or the evaluation of an even root of a negative number.

We require 4x - 3 > 0 and 2x + 3 > 0
Þ 4x > 3 and 2x > -3
Þ x > 3/4 and x > -3/2
Þ x > 3/4

At x = 3/4, f (x) = 0 - Ö (2(3/4) + 3)/2) = - Ö(9/4) = -3/2
We have a difference of two square roots, but the first root contains the greater multiple of x and will dominate the second root. The value of f (x) will increase indefinitely from its starting value of -3/2.
Therefore

The domain is [3/4, ¥)
The range is [-3/2, ¥)

Use transformations of one of the standard graphs on page 217 of the textbook,
y = | x | , y = x² , y = Ö x ,
to sketch the graph whose equation is
1. y = 2 | x - 3 |
  Starting with y = | x | :
  
  Horizontal shift 3 units to the right yields y = | x - 3 | :
  
  Vertical stretch of factor 2 completes the sketch of y = 2 | x - 3 | :
2. Starting with y = Ö x :
  
  Horizontal shift 3 units to the right yields y = Ö (x - 3) :
  
  Vertical stretch of factor 2 yields y = 2 Ö (x - 3) :
  
  Reflection in the x-axis yields y = -2 Ö (x - 3) :
  
  Vertical shift 5 units up completes the sketch of y = 5 - 2 Ö (x - 3) :
3. y = x² - 4 x + 5
  Completing the square:
  y = (x - 2)² - 4 + 5 = (x - 2)² + 1
  Starting with y = x² :
  
  Horizontal shift of 2 units to the right yields y = (x - 2)² :
  
  No vertical stretch is required.
  No reflections are required.
  A vertical shift of 1 unit up completes the sketch of y = (x - 2)² + 1 :
  
  The black curve has its vertex at (2, 1), has no x-intercept and has a y-intercept at 5.

Find the domain and inverse of the function and sketch the graph of f ^-1(x) on top of the graph of f (x):
There are no even roots present in f (x).
The only values of x that need to be excluded from the domain are those that cause the denominator to vanish.
x + 1 = 0 Þ x = -1.
Therefore the domain of f (x) is

{ x | x ¹ -1 }

Therefore the inverse function is

The graph of f ^-1 (x) (in red below) is most easily sketched by reflecting the graph of f (x) in the line y = x .

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Dr. G.H. George

MATH 1090 Algebra and Trigonometry Problem Set 4 - Solutions

MATH 1090 Algebra and Trigonometry
Problem Set 4 - Solutions