Given f (x ) = x^2 + 3x + 2 and g (x ) = x + 1, perform the indicated operations. (f − g)(x)
It does the same thing for two more but (f+g)(x) and (fg)(x) instead

(i) (f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2

Step-by-step explanation:

The given functions are;

f(x) = x² + 3·x + 2

g(x) = x + 1

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

∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1

(f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3

(f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = f(x) × g(x)

∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2

(f·g)(x) = x³ + 4·x² + 5·x + 2

ye

step-by-step explanation:

the solution is 12x-6y