What Is Fog In Math

Blended functions

Composite functions: Let A, B, C are three sets .
Let f: A -> B, k : B -> C exist two functions. Hither nosotros accept taken the domain of g to be the
co-domain of f.
g o f : A -> C equally
g o f (a) = g [f (a) ] for a ∈ A
Since f(a) ∈ B
one thousand [f(a) ] ∈ C
The function one thousand o f so obtained is called the composition of f and g.

Examples on blended functions

Case i: A = {1, ii, iii}, B = {4, 5}, C = {v, 6}
Let f : A -> B, thou: B -> C exist defined by f(1) = 4, f(2) = 5, f(3) = 4, g(4) = v, g(5) = 6. Find g o f : A -> C
Solution:We have,
f(1) = 4 and thou (iv) = 5
Then g [f(1) ] = g o f (1) = 5
f(ii) = 5 and g(f) = 6
∴ thousand [f(ii)] = g o f (two) = 6
f(three) = 4 and g(iv) = five
And then thou [f(three)] = g o f (4) = 5
And then, g o f = {(one, 5), (2, half-dozen), (3, 5)}

Example 2: Let f:R -> R be defined past f(x) = 2x – iii and k:R -> R be divers by g(x) = (x + 3)/ two. Show that f o g = g o f
Solution : f o grand is a composite-function.
f o m means m(ten) office is in f(x) part.
f o g = f[g(ten)]
f(x) = 2x -iii and k(x)= (x+3)/ii
∴ f[g(x)] =[ii(ten+3)/ii] - 3
= ten + iii - 3
f o chiliad = x -----(one)
g o f means f(x) part is in thou(x) part.
g o f = yard[f(10)]
f(ten) = 2x -3 and 1000(x) = (ten+3)/ii
g[f(x)] = [(2x-iii) + 3]/2
= (2x -3 + 3)/2
= 2x/ii
g o f = x ----(2)
And then from equation (one) and (2)
f o g = g o f

Example 3: Allow f, yard: R -> R exist defined respectively, by f(10) = x^two + 3x + i , yard(ten) = 2x -3. Observe f o g (2).
solution : f o k means m(10) role is in f(ten) function.
This means put x = 2x -3 in f(x) function.
f[1000(x)] = (2x - 3)² + three (2x -3)+ i
= (2x)² - two(2x)(three) + 3² + 6x - 9 + 1
= 4x^two - 12x + 9 + 6x - 9 +1
f[thou(x)] = f o g (x)= 4x² -6x + 1
Now we have to observe f o thousand(ii), so put ten = 2 in f o g
f o chiliad (ii)= 4(two)² - 6(2) +ane
= iv x 4 - 12 + one
= 16 -12 +1
∴ f o g(2) = 5

11th grade math
From Composite functions to Home folio

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