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frutty [35]
1 year ago
8

Let f(x)=4x-1 and g(x)=2x^2+3. Perform each function operations and then find the domain.

Mathematics
1 answer:
Triss [41]1 year ago
6 0
F(x) = 4x - 1
g(x) = 2x² + 3

1. (f + g)(x) = (4x - 1) + (2x² + 3)
    (f + g)(x) = 2x² + 4x + (-1 + 3)
    (f + g)(x) = 2x² + 4x + 2
    Domain: {x| -∞ < x < ∞}, (-∞, ∞)

2. (f - g)(x) = (4x + 1) - (2x² + 3)
    (f - g)(x) = 4x + 1 - 2x² - 3
    (f - g)(x) = -2x² + 4x + 1 - 3
    (f - g)(x) = -2x² + 4x - 2
    Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
    (g - f)(x) = 2x² + 3 - 4x + 1
    (g - f)(x) = 2x² - 4x + 3 + 1
    (g - f)(x) = 2x² - 4x + 4
    Domain: {x| -∞ < x < ∞}, (-∞, ∞)

4. (f · g)(x) = (4x + 1)(2x² + 3)
    (f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
    (f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
    (f · g)(x) = 8x³ + 12x + 2x² + 3
    (f · g)(x) = 8x³ + 2x² + 12x + 3
    Domain: {x| -∞ < x < ∞}, (-∞, ∞)

5. (\frac{f}{g})(x) = \frac{4x - 1}{2x^{2} + 3}
    Domain: 2x² + 3 ≠ 0
                         - 3  - 3
                        2x² ≠ 0
                         2      2
                          x² ≠ 0
                           x ≠ 0
                  (-∞, 0) ∨ (0, ∞)

6. (\frac{g}{f})(x) = \frac{2x^{2} + 3}{4x - 1}
    Domain: 4x - 1 ≠ 0
                      + 1 + 1
                        4x ≠ 0
                         4     4
                         x ≠ 0
                (-∞, 0) ∨ (0, ∞)
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VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0
1 year ago
If Line segment C B. bisects ∠ACD, what additional information could be used to prove ΔABC ≅ ΔDBC using SAS? Select three option
meriva

Answer:

Option (1)

Step-by-step explanation:

In the figure attached,

BC is the angle bisector of angle ACD.

To prove ΔABC and ΔDBC congruent by SAS property we require two sides and the angle between these sides to be congruent.

Since BC ≅ BC [Reflexive property]

∠ABC ≅ ∠CBD ≅ 125°

And sides AB ≅ BD

Both the triangles will be congruent.

Therefore, additional information required to prove ΔABC ≅ ΔDBC have been given in option (1).

Therefore, Option (1) will be the answer.

3 0
1 year ago
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Efectuati impartirea 12xy:3x
adelina 88 [10]

In this question , we have to simplify the given ratio, which is

12xy:3x

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We can write 12 as 3 times 4, that is

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So the common factor is 3x.

In the next step, we cancel out 3x

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Damm [24]

Complete question:

The line graph relating to the question was not attached. However, the line graph has can be found in the attachment below.

Answer:

17,209

Step-by-step explanation:

The line graph provides information about alcohol-related highway fatalities between year 2001 to 2010.

Determine the average number of alcohol-related fatalities from 2001 to 2006. Round to the nearest whole number.

The average number of alcohol related fatalities between 2001 - 2006 can be calculated thus :

From the graph:

Year - - - - - - - - - - Number of fatalities

2001 - - - - - - - - - - 17401

2002 - - - - - - - - -  17525

2003 - - - - - - - - -  17013

2004 - - - - - - - - - 16694

2005 - - - - - - - - - 16885

2006 - - - - - - - - - 17738

To get the average :

Sum of fatalities / number of years

(17401 + 17525 + 17013 + 16694 + 16885 + 17738) / 6

= 103256 / 6

= 17209.333

Average number of alcohol related fatalities is 17,209 (to the nearest whole number)

6 0
1 year ago
If -3y+5=-4 find the value of 5y
artcher [175]
-3y + 5 = -4   Subtract 5 from both sides
      -3y = -9   Divide both sides by -3
         y = 3

Now, plug that into 5y

 5y    Plug in 3
5(3)   Multiply
 15

5y = 15
6 0
1 year ago
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