Answers on Engineering Math

Answers on Engineering Math

1. The interval of the following inequalities

a. The interval notation can be use to describe the solution set of inequality. This solution for inequality in the value of x is the number that substitutes to the number x so that it can have the true statement. The solution therefore for the example is given by:

2x + 5 <  x + 8

3x – x < 8 – 5

2x < 3

x < 3/2

This implies that the interval in the given inequalities is (-inf , 3/2) or all the values less than or equal to 3/2. In number line we have:

For the graph we have:

 

 

b. The equation   2 – x/2   < ½ will can be solve simultaneously by multiplying both sides of the equation by 2. Then we have:

  4 – x     < 1

This can lead to the question which number for the given equation are less than a unit apart from 4 and will satisfy the given equation. This can yield to the interval of (<3 and >5). The interval had been given below:

2. The point of intersection of the curves y – 3x = 2 or y = 3x +2 and y = x2 -3x + 4

The first equation is a line y = 3x + 2

The second equation is a parabola y = x2 – 3x + 4

The points of intersection of the two functions are their point of intersection.  

The coordinates can be calculated in this way:

y = 3x + 2

y = x2 – 3x + 4

To set the two equations in terms of y-coordinates will be like:

3x + 2 = x2 – 3x + 4

0 = x2 – 6x + 2

With the use of the quadratic formula:

x = 6 + √ 36 – 4(1)(2)

                 2(1)

x = 6 +√28 / 2            and                 x = 6 – √28 / 2  

x = 5.65                                              x = 0.35

Substituting these values to the given equation will yield:

If x = 5.65

y = 3 (5.65) + 2

y = 18.95

y = x2 – 3x + 4

y = 5.652 – 3(5.65) + 4

y = 18.95

 

If x  = 0.35

y =  3x + 2

y = 3(0.35) + 2

y = 3.05

 

y = x2 – 3x + 4

y = (0.35)2 – 3(0.35) + 4

y = 3.05

(The answers are a little off bit due to the truncating and rounding of the decimals along the way; I decided to use the same value to make it uniform)

The graph of the graphs and their point of intersections are given below:

3. a). The domain of the functions y = f(x-1) are all the values of f(x) as varies in x. Therefore the domain for the given equation is all values of x or the set of all real numbers. Though graphically, the domain will be (0, 1) and the range of (0, -1) or all values of y or the set of all real numbers.

 

 

 

b. The domain for the function y = 2 – f (x – 3) or it can also be y = – x – 1 are all the set of real numbers for the value of x while its range are also the set of real number for the value of y. The graph of the function is given below and has the domain of (-1, 0) and range of (0, -1).

 

4. a).  z2 – 2iz + 3 = 0

Using the quadratic formula:

        =   2i + √(-2 I ) – 4(1) (3)

                            2(1)

        =  2i + √-16

     2(1)

       z =  2i  + 4 i                     z =  2i  - 4 i

                 2(1)                                  2(1)

       z = 3 i                   and      z = – i

   

 

  

b. ) The cube root of i  is z

then, z3 + i = 0 which is (z + i ) (z2  - z + i)

Then, the cube roots are – i, and using the quadratic formula we have:

= 1 + √ 1- 4(1) i 

            2(1)

The other roots are:

1+√ – 3i        and  1 – √ -3i

    2                              2

 

5. a).  Expressing the value of cos (3x) in terms of sin(x) and cost (x)

            cos (3x) = cos(2x + x) = cos (2x) cos (x) – sin(2x) sin (x)

            Through eliminating the functions of 2x, use the cos (2x) = 2 cos2x – 1 and sin (2x) = 2sinx cos x.

            cos (2x) cos (x) – sin(2x)sin (x) = (2cos2x – 1) cos x – 2 sinx cos x sin x

Through simplification of this can give the final answer in terms of cos(x) and sin(x)

2cos3 x – cos x – 2 sin2x cos x 

 

 

b.) The simplified value of 1 – cos x / sin x can be in the ratio of tan (x/2).

To prove:

1 – cos (x) / sin (x)

Using the double angles identities,

cos (2y) = cos2y– sin2y

sin (2y) = 2sin(y) cos(y)

Therefore, letting the value of y = x/2, and get,

cos 2 x/2 = cos2(x/2)– sin2(x/2), or it may be:

cos(x) = cos2(x/2)- sin2(x/2)

In similar term:

sin (x) = 2 sin (x/2) cos (x/2)

Simplify further,

[1 – {cos2(x/2) + sin2 (x/2)}] / [2sin (x/2) cos (x/2)]

Though 1 – cos 2 (x/2) = sin2 (x/2)

   [2 sin2 (x/2) / 2sin (x/2) cos (x/2)]

Cancellation for the 2 and single value of sin(x/2) will get:

[sin(x/2) / cos (x/2)]

This is also equal to

tan (x/2)

 

 

6. y = (3√x – 2/x) (4√x +1/x)

    To make it  simpler, get the LCD for both factors as:

    y = (3x√x – 2 / x) (4x√x +1 / x)  or

    y = (3×3/2 – 2 / x) (4×3/2 + 1)

    Multiplying the both equations we have:

    y = 12×3 – 5×3/2 – 2

                        x2

   Using the quotient rule:

y’ = (x2) (D) (12×3 – 5×3/2 – 2) – (12×3 – 5×3/2 – 2) (D) (x2)

                                                (x2)2

y’ = (x2) (36×2 – 15/2×1/2 ) – (12×3 – 5×3/2 – 2) (2x)

                                       (x4)

y’ = 36×4 – 15/2×5/2 – (24×4 – 10×5/2 – 4x)

                                    (x4)

y’ = 36×4 – 15/2×5/2 – 24×4 + 10×5/2 – 4x

                                    x4

y’ = 12×4 +15/2×5/2 + 4x

                  x4

y’ = 12 + 5/2×3/2  + 4/x3

y’ =  4 +   5    + 12

      x3    2x√x         .

 

 

 

b. y = 1 + x2 / 2 – x2

    Using the quotient rule:

y = (2-x2) (D) (1 + x2) – (1 + x2) (D) (2-x2)

                                    (2-x2)2

y = (2-x2) (2x) – (1 + x2) (-2x)

                    (2-x2)2

y = 4x – 2×3 – (-2x – 2×3)

                   (2-x2)2

y = 4x – 2×3 + 2x + 2×3)

                   (2-x2)2

y =        6x          .

       4 – 4×2 + x4

y = 3x  -  1   +   1

       2     4x       x2

 

y = 1  -    1   +  3x

      x4 +     4x       2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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