Mathematics, 15.11.2019 12:31 rol15

Match each function with the corresponding restrictions to its domain when p(x)=2x and q(x) = 2^x.

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Recall that the domain of a function is the set of all x values of the function for which the function is defined.

Given that
p(x)=2x

and

Given that
$r(x)= \frac{q(x)}{p(x)} = \frac{2^x}{2x}$
For the function to be undefined, the denominator should be zero
i.e. 2x = 0 which means that x = 0
Thus, the domain of the function is all real values of x except 0.

Therefore, the restriction is x ≠ 0.

Given that
$r(x)= \frac{1}{p(x)-q(x)} = \frac{1}{2x-2^x}$
For the function to be undefined, the denominator should be zero
i.e.
$2x-2^x=0 \\ 2^x=2x$
For x = 2,
2^2 = 2(2)=4

Also, for x = 1
2^1 = 2(1)=2

Thus, the domain of the function is all real values of x except 1 and 2.

Therefore, the restriction is x ≠ 1, 2.

Given that
$r(x)= \frac{1}{p(x)+16q(x)}=\frac{1}{2x+16(2^x)}$
For the function to be undefined, the denominator should be zero
i.e.
$2x+16(2^x)=0 \\ 2x+2^4(2^x)=0 \\ 2x+2^{4+x}=0$
For x = -2,
$2(-2)+2^{4-2}=-4+2^2=-4+4=0$ .
Thus, the domain of the function is all real values of x except -2.

Therefore, the restriction is x ≠ -2.

Given that
$r(x)=p(x)\circ q(x)=p(q(x))=2(2^x)$
There is no value of x, for which the function is undefined.
Thus, the domain of the function is all real values of x.

Therefore, there is no restrictions to the domain of the function.

Given that
$r(x)= \frac{q(x)}{p(x)+2}=\frac{2^x}{2x+2}$
For the function to be undefined, the denominator should be zero
i.e.
$2x+2=0 \\ 2x=-2 \\ x = -1$
Thus, the domain of the function is all real values of x except -1.

Therefore, the restriction is x ≠ -1.

Answer from: Quest

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

the function given is y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 = -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24

Answer from: Quest

step-by-step explanation:

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Match each function with the corresponding restrictions to its domain when p(x)=2x and q(x) = 2^x....

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