Problem 1 :
If f, g : R → R are defined by f(x) = |x| + x and g(x) = |x| − x, find g ◦ f and f ◦ g.
Solution :
f(x) = |x| + x = x + x (or) -x + x f(x) = 2x (or) f(x) = 0 |
g(x) = |x| - x = x - x (or) -x - x g(x) = 0 (or) g(x) = -2x |
If f(x) = 2x and g(x) = 0 gof = g [f(x)] = g [f(x)] = g[2x] gof = 0 If f(x) = 0 and g(x) = -2x gof = g [f(x)] = g [0] gof = 0 |
If f(x) = 2x and g(x) = 0 fog = f [g(x)] = f [g(x)] = f[0] fog = 0 If f(x) = 0 and g(x) = -2x fog = f [g(x)] = f [-2x] fog = 0 |
For all values of x, fog and gof is 0.
Problem 2 :
If f, g, h are real valued functions defined on R, then prove that (f + g) ◦ h = f◦h + g ◦ h. What can you say about f ◦ (g + h) ? Justify your answer.
Solution :
(f + g) ◦ h = (f + g) [h(x)]
= f [h(x)] + g [h(x)]
= f◦h + g ◦ h ----(1)
f ◦ (g + h) = f [(g + h)(x)]
= f[g(x) + h(x)]
= f[g(x)] + f[h(x)]
= f ◦ g + f ◦ h ----(2)
It is a function.
Problem 3 :
If f : R → R is defined by f(x) = 3x − 5, prove that f is a bijection and find its inverse.
Solution :
f(x) = 3x − 5
f : A → B is said to be bijective if it is both one-to-one and onto.
For all x ∊ A, we get different values of y ∊ B, so it is one to one function.
For all y ∊ B, there is a preimage in A. So it is onto function.
Hence it is bijective function.
Inverse function :
Let y = 3x - 5
3x = y + 5
x = (y + 5)/3
f^{-1}(x) = (x + 5)/3
Problem 4 :
The weight of the muscles of a man is a function of his body weight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.
Solution :
Given that :
W(x) = 0.35x
Here "x" represents weight of the body, it will not be negative. Hence its domain will be > 0.
Problem 5 :
The distance of an object falling is a function of time t and can be expressed as s(t) = −16t^{2}. Graph the function and determine if it is one-to-one.
Solution :
From the above graph, we come to know that for different values of t ∊ A, we get different value of s(t) ∊ B. Hence it is one to one function.
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