wyznacz funkcje liniową f, aby dla dowolnego x rzeczywistego zachodził warunek: F(3x-5) = 4x-4
x = 1
Mamy
f( 3*1 - 5) = 4*1 - 4
czyli
f(-2) = 0
=========
x = 2
f( 3*2 - 5) = 4*2 - 4
f( 1) = 4
Mamy punkty : A = ( -2; 0) , B = ( 1; 4)
f(x) = a x + b
0 = -2 a + b
4 = a + b
----------------- odejmujemy stronami
4 - 0 = a - ( -2a)
4 = 3 a / : 3
a = 4/3
=======
b = 4 - a = 4 - 4/3 = 12/3 - 4/3 = 8/3
===================================
Odp. f(x) = ( 4/3) x + 8/3
==========================
spr.
f( 3x - 5) = (4/3)*( 3x - 5) + 8/3 = 4x - 20/3 + 8/3 = 4x - 12/3 = 4x - 4
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
x = 1
Mamy
f( 3*1 - 5) = 4*1 - 4
czyli
f(-2) = 0
=========
x = 2
f( 3*2 - 5) = 4*2 - 4
czyli
f( 1) = 4
Mamy punkty : A = ( -2; 0) , B = ( 1; 4)
f(x) = a x + b
czyli
0 = -2 a + b
4 = a + b
----------------- odejmujemy stronami
4 - 0 = a - ( -2a)
4 = 3 a / : 3
a = 4/3
=======
b = 4 - a = 4 - 4/3 = 12/3 - 4/3 = 8/3
===================================
Odp. f(x) = ( 4/3) x + 8/3
==========================
spr.
f( 3x - 5) = (4/3)*( 3x - 5) + 8/3 = 4x - 20/3 + 8/3 = 4x - 12/3 = 4x - 4