MAT222 assignment 5
Composition and Inverse
Chadwick A. Reynolds
MAT222: Intermediate Algebra
P. Martin Fenlon
September 21, 2014
Composition and Inverse
The functions that will be used in this week’s assignment will insert the x’s and y’s that will substitute their selves when the function is inversed, basically they will have the same points. The functions will be plugged in place of “x” for the function composition. Functions can be used in different things that are in our lives, such as vending machines, factory machines, thermometers, etc. I will add these functions together or look for the opposite, those are called the composition of functions and inverse of a function, respectively. These are my following functions problem here.
1. f(x) = 2x +5
2. G(x) = x2 - 3
3. h(x) = 7 – x 3 I first had to compute f - h4. Once the solution is rewritten is becomes easier as it is rewritten as a different format. I will then solve the problem in this way: f - h(4) Original expression.
(f – h)(4) = f(4) – h(4) f(4) = 2(4) + 5
= 8 + 5 = 13 f(4) = 13 h(4) = 7 - 4 3 = 3/3 = 1 h(4) = 1
(f – h)(4) = 13 – 1 = 12 The second problem involves composing two pairs into each other. f°g(x)- Original expression
f(g(x)) Rewritten so that rule of f will apply
fx2 - 3 g inserted
2x2- 3 + 5 f applied to g
2x2 - 6 + 5 Simplified
2x2 - 1 Solution
h (g(x)) Rewritten so that rule of h will apply to g
hx2-3
Chadwick A. Reynolds
MAT222: Intermediate Algebra
P. Martin Fenlon
September 21, 2014
Composition and Inverse
The functions that will be used in this week’s assignment will insert the x’s and y’s that will substitute their selves when the function is inversed, basically they will have the same points. The functions will be plugged in place of “x” for the function composition. Functions can be used in different things that are in our lives, such as vending machines, factory machines, thermometers, etc. I will add these functions together or look for the opposite, those are called the composition of functions and inverse of a function, respectively. These are my following functions problem here.
1. f(x) = 2x +5
2. G(x) = x2 - 3
3. h(x) = 7 – x 3 I first had to compute f - h4. Once the solution is rewritten is becomes easier as it is rewritten as a different format. I will then solve the problem in this way: f - h(4) Original expression.
(f – h)(4) = f(4) – h(4) f(4) = 2(4) + 5
= 8 + 5 = 13 f(4) = 13 h(4) = 7 - 4 3 = 3/3 = 1 h(4) = 1
(f – h)(4) = 13 – 1 = 12 The second problem involves composing two pairs into each other. f°g(x)- Original expression
f(g(x)) Rewritten so that rule of f will apply
fx2 - 3 g inserted
2x2- 3 + 5 f applied to g
2x2 - 6 + 5 Simplified
2x2 - 1 Solution
h (g(x)) Rewritten so that rule of h will apply to g
hx2-3