Question

if f, g:R»R are defined respectively by f(x) =x2+3x+1 and g(x)=2x—3,then find the value of f pg(x)​

Answer 1

Given that,f(x) = x2 + 3x + 1,g(x) = 2x - 3

(i) fog = f(g(x))

= f(2x-3)

= (2x-3)2 + 3(2x-3)+1

= 4x2 - 12x + 9 + 6x - 9 + 1

= 4x2 - 6x + 1

(ii) gof = g(f(x)

= g(x2+3x+1)

= 2(x2+3x+1)-3

= 2x2 + 6x - 1

(iii) fof = f(f(x))

= f(x2 + 3x + 1)

= (x2 + 3x + 1)2 + 3(x2 + 3x + 1) + 1

= x4 + 9x2 + 1 + 6x2 + 6x + 2x2 + 3x2 + 9x + 3 + 1

= x4 + 6x3 + 14x2 + 15x + 5

(iv) gog = g(g(x))

= g(2x-3)

= 2(2x-3)-3

= 4x - 6 - 3

= 4x - 9

Answer 2

Answer:

Step-by-step explanation:

$f(g(x))=(g(x))^{2} +3g(x)+1=(2x-3)^{2} +3(2x-3)+1=4x^{2} -12x+9+6x-6+1\\=4x^{2} -6x+4$