Найдите число целых значений функции
y=2^(1-4x-x^2)

Предыдущий
вопрос Следующий
вопрос

Для нахождения целых значений функции we need to consider the possible values of the exponent (1-4x-x^2) since 2 raised to any real number will result in a real number.

The exponent (1-4x-x^2) can be any real number, but for the purpose of finding integer values, we need it to be an integer.

If (1-4x-x^2) is an integer, then the function y=2^(1-4x-x^2) will also be an integer.

So, the possible integer values for the function y=2^(1-4x-x^2) will depend on the different integer values that (1-4x-x^2) can take.

We need to solve the quadratic equation x^2 + 4x + 1 = 1, in order to find the possible integer values of x.

From the quadratic equation x^2 + 4x + 1 = 1, we have:
x^2 + 4x = 0
x(x+4) = 0

Therefore, the possible integer values for x are x = 0 and x = -4.

For x = 0:
y = 2^(1-4(0) - (0)^2) = 2^1 = 2

For x = -4:
y = 2^(1-4(-4) - (-4)^2) = 2^(1+16-16) = 2

Therefore, the function y=2^(1-4x-x^2) has only one integer value which is y = 2.