Дана функция y = f(x), где f(x) = 2x + 3. Найдите:
а) f(−2), f(−0,5), f(0), f(1,5);
б) $f(-p), f(\frac{p}{2}), f(0,5 + p), f(p) + 0,5$;
в) $f(y^2), f(y^2 + 2), f((y + 2)^2), f(y^2) + 2$;
г) $f(x - 4), f(1 - x), f(2x^2) - 4, f(\frac{1}{2}x^3 - 1)$.
f(−2) = 2 * (−2) + 3 = −4 + 3 = −1
f(−0,5) = 2 * (−0,5) + 3 = −1 + 3 = 2
f(0) = 2 * 0 + 3 = 0 + 3 = 3
f(1,5) = 2 * 1,5 + 3 = 3 + 3 = 6
f(−p) = 2 * (−p) + 3 = −2p + 3
$f(\frac{p}{2}) = 2 * \frac{p}{2} + 3 = p + 3$
f(0,5 + p) = 2(0,5 + p) + 3 = 1 + 2p + 3 = 2p + 4
f(p) + 0,5 = 2p + 0,5 + 3 = 2p + 3,5
$f(y^2) = 2y^2 + 3$
$f(y^2 + 2) = 2(y^2 + 2) + 3 = 2y^2 + 4 + 3 = 2y^2 + 7$
$f((y + 2)^2) = 2(y + 2)^2 + 3 = 2(y^2 + 4y + 4) + 3 = 2y^2 + 8y + 8 + 3 = 2y^2 + 8y + 11$
$f(y^2) + 2 = 2y^2 + 3 + 2 = 2y^2 + 5$
f(x − 4) = 2(x − 4) + 3 = 2x − 8 + 3 = 2x − 5
f(1 − x) = 2(1 − x) + 3 = 2 − 2x + 3 = 5 − 2x
$f(2x^2) - 4 = 2 * 2x^2 - 4 + 3 = 4x^2 - 1$
$f(\frac{1}{2}x^3 - 1) = 2(\frac{1}{2}x^3 - 1) + 3 = x^3 - 2 + 3 = x^3 + 1$
Пожауйста, оцените решение
