If you’re trying to find f(-1) it would be -20
Answer:
(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
Given

We have to set the restraint

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.
Similarly, we have

So, we have to impose

Squaring both sides, we have

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.
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