Дана функция y = f(x), где f(x) = 4x − 1. Найдите:
а) $f(-3), f(0), f(0,5), f(\frac{1}{4})$;
б) f(a), f(−2a), f(a − 2), f(a) − 2;
в) $f(t^2), f(t^2 - 1), f((t - 1)^2), f(t^2) - 1$;
г) $f(x + 3), f(2x - 1), f(1 - 2x)^2, f(x - x^2)$.
f(−3) = 4 * (−3) − 1 = −12 − 1 = −13
f(0) = 4 * 0 − 1 = 0 − 1 = −1
f(0,5) = 4 * 0,5 − 1 = 2 − 1 = 1
$f(\frac{1}{4}) = 4 * \frac{1}{4} - 1 = 1 - 1 = 0$
f(a) = 4a − 1
f(−2a) = 4 * (−2a) − 1 = −8a − 1
f(a − 2) = 4 * (a − 2) − 1 = 4a − 8 − 1 = 4a − 9
f(a) − 2 = 4a − 2 − 1 = 4a − 3
$f(t^2) = 4t^2 - 1$
$f(t^2 - 1) = 4(t^2 - 1) - 1 = 4t^2 - 4 - 1 = 4t^2 - 5$
$f((t - 1)^2) = 4(t - 1)^2 - 1 = 4(t^2 - 2t + 1) - 1 = 4t^2 - 8t + 4 - 1 = 4t^2 - 8t + 3$
$f(t^2) - 1 = 4t^2 - 1 - 1 = 4t^2 - 2$
f(x + 3) = 4(x + 3) − 1 = 4x + 12 − 1 = 4x + 11
f(2x − 1) = 4(2x − 1) − 1 = 8x − 4 − 1 = 8x − 5
$f(1 - 2x)^2 = 4(1 - 2x)^2 - 1 = 4(1 - 4x + 4x^2) - 1 = 4 - 16x + 16x^2 - 1 = 16x^2 - 16x + 3$
$f(x - x^2) = 4(x - x^2) - 1 = 4x - 4x^2 - 1 = -(4x^2 - 4x + 1) = -(2x - 1)^2$
Пожауйста, оцените решение
