Question

If f(x) = 3x + 5, g (x) = 6x - 1, then find
(a) (f+g) (x)
(b) (f-g) (2)
(c) (tg) (3) (d) f/g)(x) and its domain​

Answer 1

Step-by-step explanation:

(a) (f+g) x = F(x) + g(x)

= 9x + 4

(b) (f-g) 2 = f(2) + g(2)

= 3(2) + 5 + 6(2) -1

= 22

(d) f(x) / g(x) = 3x + 5 / 6x -1

For Domain ,

6x - 1 cannot be equal to zero

6x cannot be equal to 1

x cannot be equal to 1/6

Domain = Set of all real no. except 1/6

Answer 2

a) (f + g)(x) = 9x + 4, (b) (f - g)(2) = 0, (c) (fg)(x) $= 18x^2+27x-5$

(d) $(\dfrac{f}{g} )$(x) $=\dfrac{3x+5}{6x-1}$ and Domain = Set of all real numbers except $\dfrac{1}{6}$.

Step-by-step explanation:

Given,

f(x) = 3x + 5, g (x) = 6x - 1

To find, (a) (f + g)(x)  (b) (f-g) (2)  (c) (tg) (3) (d) $(\dfrac{f}{g} )$(x) and its domain​.

a) (f + g)(x) = f(x) + g(x)

= 3x + 5 + 6x - 1

= 9x + 4

(b) (f - g)(2)

(f - g)(x) = f(x) - g(x)

= 3x + 5 - 6x + 1

= - 3x + 6

∴ f(2) + g(2)

= - 3(2) + 6

= - 6 + 6

= 0

(c) (fg)(x) = f(x).g(x)

= (3x + 5)(6x - 1)

$= 18x^2-3x+30x-5$

$= 18x^2+27x-5$

(d) $(\dfrac{f}{g})(x)=\dfrac{f(x)}{g(x)}$

$=\dfrac{3x+5}{6x-1}$

For domain,

6x - 1 ≠ 0

6x ≠ 1

x ≠  $\dfrac{1}{6}$

Domain = Set of all real numbers except $\dfrac{1}{6}$.