Let f (x) = x ^ 2 + 2, g (x) = x + a, determine the value of "a" so that (fog) (3) = (gof) (a-1).
If g (x) = 1, 0 ≤ x ≤ 1
2, 1
-
-
If g (x) = 1, 0 ≤ x ≤ 1
2, 1
-
(f0g)(3) = f(g(3)) = f(3+a) = (3+a)^2 + 2 = 11 + 6a + a^2
(gof)(a-1) = g(f(a-1)) = g((a-1)^2 + 2) = (a-1)^2 + 2 +a = a^2 -a +3
-------------------
(fog)(3) = (gof)(a-1)
a^2 + 6a + 11 = a^2 -a +3
6a + 11 = -a + 3
7a = -8
a = -8/7
(gof)(a-1) = g(f(a-1)) = g((a-1)^2 + 2) = (a-1)^2 + 2 +a = a^2 -a +3
-------------------
(fog)(3) = (gof)(a-1)
a^2 + 6a + 11 = a^2 -a +3
6a + 11 = -a + 3
7a = -8
a = -8/7
-
(fog) (3) = (gof) (a - 1)
= f(3 + a) = g(a^2 - 2a + 3)
a^2 + 6a + 11 = a^2 - a + 3
7a = -8 => a = -8/7
= f(3 + a) = g(a^2 - 2a + 3)
a^2 + 6a + 11 = a^2 - a + 3
7a = -8 => a = -8/7
1
keywords: problems,Help,math,Pleasee,me,Pleasee!! Help me!! :( math problems