The domain and target set of functions f and g isR. The functions are definedas:(b)•f(x) = 2x+ 3•g(x) = 5x+ 7(a)f◦g?(b)g◦f?(c) (f◦g)−1?(d)f−1◦g−1?(e)g−1◦f−1?

Step-by-step explanation:

Given the domain and target set of functions f and g expressed as;

f(x) = 2x+3 an g(x) = 5x+7 we are to find the following;

a) f◦g

f◦g = f[g(x)]

f[g(x)] = f[5x+7]

To get f(5x+7), we will replace the variable x in f(x) with 5x+7 as shown;

f(x) = 2x+3

f(5x+7) = 2(5x+7)+3

f(5x+7) = 10x+14+3

f(5x+7) = 10x+17

Hence f◦g = 10x+17

b) g◦f

g◦f = g[f(x)]

g[f(x)] = g[2x+3]

To get g(2x+3), we will replace the variable x in g(x) with 2x+3 as shown;

g(x) = 5x+7

g(2x+3) = 5(2x+3)+7

g(2x+3) = 10x+15+7

g(2x+3) = 10x+22

Hence g◦f = 10x+22

c) For (f◦g)−1 (inverse of (f◦g))

Given (f◦g) = 10x+17

To find the inverse, first we will replace (f◦g) with variable y to have;

y = 10x+17

Then we will interchange variable y for x:

x = 10y+17

We will then make y the subject of the formula;

10y = x-17

y = x-17/10

Hence the inverse of the function

(f◦g)−1 = (x-17)/10

d) For the function f−1◦g−1

We need to get the inverse of function f(x) and g(x) first.

For f-1(x):

Given f(x)= 2x+3

To find the inverse, first we will replace f(x) with variable y to have;

y = 2x+3

Then we will interchange variable y for x:

x = 2y+3

We will then make y the subject of the formula;

2y = x-3

y = x-3/2

Hence the inverse of the function

f-1(x) = (x-3)/2

For g-1(x):

Given g(x)= 5x+7

To find the inverse, first we will replace g(x) with variable y to have;

y = 5x+7

Then we will interchange variable y for x:

x = 5y+7

We will then make y the subject of the formula;

5y = x-7

y = x-7/5

Hence the inverse of the function

g-1(x) = (x-7)/5

Now to get )f−1◦g−1

f−1◦g−1 = f-1[g-1(x)]

f-1[g-1(x)] = f-1(x-7/5)

Since f-1(x) = x-3/2

f-1(x-7/5) = [(x-7/5)-3]/2

= [(x-7)-15/5]/2

= [(x-7-15)/5]/2

= [x-22/5]/2

= (x-22)/10

Hence f−1◦g−1 = (x-22)/10

e) For the composite function g−1◦f−1

g−1◦f−1 = g-1[f-1(x)]

g-1[f-1(x)] = g-1(x-3/2)

Since g-1(x) = x-7/5

g-1(x-3/2) = [(x-3/2)-7]/5

= [(x-3)-14)/2]/5

= [(x-17)/2]/5

= x-17/10

Hence g-1◦f-1 = (x-17)/10

No, it is irrational.

Step-by-step explanation:

The quotient of √10 and 5 is equal to which equals 0.6245553...

The number goes on and on forever with no determinable pattern (i.e. the decimal 0.234234234... although the decimal goes on forever, the 234 repeats over and over).

If you're still confused, check out https://www.mathsisfun.com/irrational-numbers.html I find it to be a really helpful website!