Question

Для функции f(х)=х2-5 сравните указанные значения функции 1. f(-1/2) и f(0,5)2. f(0) и f(-2)3. f(-2) и f(1)4. f(-3) и f(4)

Answer 1

Решение задания смотри на фотографии

Answer 2

To compare the specified values for the function f(x) = x^2 - 5, let's calculate each expression:

1. f(-1/2) and f(0.5): f(-1/2) = (-1/2)^2 - 5 = 1/4 - 5 = -19/4 ≈ -4.75 f(0.5) = (0.5)^2 - 5 = 1/4 - 5 = -19/4 ≈ -4.75

   Both f(-1/2) and f(0.5) are approximately equal to -4.75.

2. f(0) and f(-2): f(0) = (0)^2 - 5 = 0 - 5 = -5 f(-2) = (-2)^2 - 5 = 4 - 5 = -1

   f(0) is -5, and f(-2) is -1. Therefore, f(0) is less than f(-2).

3. f(-2) and f(1): f(-2) = (-2)^2 - 5 = 4 - 5 = -1 f(1) = (1)^2 - 5 = 1 - 5 = -4

   f(-2) is -1, and f(1) is -4. Therefore, f(1) is less than f(-2).

4. f(-3) and f(4): f(-3) = (-3)^2 - 5 = 9 - 5 = 4 f(4) = (4)^2 - 5 = 16 - 5 = 11

   f(-3) is 4, and f(4) is 11. Therefore, f(4) is greater than f(-3).

So, the comparisons are:

1. f(-1/2) ≈ f(0.5)
2. f(0) < f(-2)
3. f(1) < f(-2)
4. f(4) > f(-3)