Answer:
(f + g)(x) = I2x + 1I + 1 ⇒ C
Step-by-step explanation:
∵ f(x) = I2x + 1I + 3
∵ g(x) = -2
→ We need to find (f + g)(x), which means add the two functions
∵ (f + g)(x) = f(x) + g(x)
→ Substitute the right side of each function on the right side
∴ (f + g)(x) = I2x + 1l + 3 + (-2)
→ Remember (+)(-) = (-)
∴ (f + g)(x) = I2x + 1I + 3 - 2
→ Add the like terms in the right side
∵ (f + g)(x) = I2x + 1I + (3 - 2)
∴ (f + g)(x) = I2x + 1I + 1
i hope this work for you
