MATH120 (2014-1)

TUTORIAL 2 - SOLUTIONS

Question 1. Solve the following equations for x.

(a) 7x + 15 = 6;

7x = 6− 15 = −9,

x = −97.

(b) x−23 = 3;(

x−23

)3

= 33,

x−2 = 27,1x2 = 27,

x2 = 127

,

x = ± 13√

3.

(c) |x| = 3;

x = ±3.

(d) |x− 2| = 3;

x− 2 = ±3

x− 2 = 3 or x− 2 = −3

x = 3 + 2 = 5 or x = −3 + 2 = −1.

There are two solutions: x = 5 and x = −1.

Question 2. Find all real numbers, x, which satisfy the following inequalities.

(a) |x| ≤ 3;

Geometrically it means that the distance between x and origin is smaller or equal

to 3, see the picture below.

−3 ≤ x ≤ 3.

(f) |3− x| > 20

The distance between x and 3 is larger than 20, see the picture below.

It is true for x > 23 and for x < −17.

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Question 2’. Sketch the graphs of the function given by the formula

(a) y = −2x2 + 6x + 3 (R −→ R);

(b) y = x2 + x + 1 ( R −→ R).

In each case find the maximum or minimum value of the function and its zeros,

that is, all x for which y = 0.

(a) Solution The graph of a quadratic function ax2 + bx + c is a parabola.

As the coefficient at x2 is negative (a = −2), the parabola is concave up. The

coordinates of the vertex are given by(− b

2a,− b2

4a+ c)

. For a = −2, b = 6, and

c = 3 we find (32, 71

2). It is a maximum, because the parabola is concave down.

The parabola intersects with the y-axis at x = 0: y = −2 × 02 + 6 × 0 + 3 = 3.

The parabola intersects the x axis if −2x2 + 6x + 3 = 0. We solve this quadratic

equation:

x1,2 =−b±

√b2 − 4ac

2a=−6±

√36 + 24

−4=−6±

√60

−4=

3±√

15

2.

See the sketch on the picture below.

(b) Solution

As a = 1 is positive, this parabola is concave up, and the vertex is a minimum.

Its coordinates are(− b

2a,− b2

4a+ c)

= (−12, 3

4). The parabola intersects with the

y-axis at x = 0: y = 02 + 0 + 1 = 1. It does not intersect with the x-axis,

as the equation x2 + x + 1 = 0 has no solutions in real numbers. Note that

3

x2 + x + 1 = (x + 12)2 + 3

4. We get the sketch of the parabola by moving the graph

of y = x2 by 12

units to the left and 34

units upward.

See the sketch in the picture below

Question 3. Consider the two curves given by

y = x + 2 and y = x2 − 2x + 1.

Decide whether the two curves intersect. If they do intersect, find their points of

intersection.

Solution. We solve the two equations simultaneously: x + 2 = x2 − 2x + 1, or

x + 2 = x2 − 3x− 1 = 0.

4

x1,2 =3±

√32 − 4× 1× (−1)

2× 1=

3±√

13

2.

We find the corresponding y-coordinates from the first equation y = x + 2:

y1 = 3+√

132

+ 2 = 7+√

132

and y2 = 3−√

132

+ 2 = 7−√

132

.

There are two intersection points: (3+√

132

, 7+√

132

) and (3−√

132

, 7−√

132

).

Question 4.

(a) Sketch the graph of the function given by the formula y = x3 (R −→ R);

(b) Using the vertical or horizontal translations of the graph y = x3, sketch the

graphs of the functions y = x3 − 2, y = (x− 1)3, and y = (x− 1)3 + 3.

Solution. (a) See the picture below.

(b) (i) We move the graph of y = x3 two units downwards and get the graph of

y = x3 − 2.

(ii) Move the graph of y = x3 one unit to the right and get the graph of

y = (x− 1)3.

(iii) Move the graph of the function y = (x − 1)3 three units upwards and get

the graph of y = (x− 1)3 + 3.

Question 5. Find the intersection points of the graphs of the functions f(x) =

|x + 5| and g(x) = 10.

Solution.

Solve the two given equations simultaneously: |x + 5| = 10, x + 5 = ±10,

x1 = 10− 5 = 5 and x2 = −10− 5 = −15.

There are two intersection points: (5, 10) and (−15, 10).

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