Answer:
[tex](f + g)(x) = 7 - 6x[/tex] --- (1)
[tex](f - g)(x) = 6x - 1[/tex] -- (2)
[tex](f . g)(x) =4(5^x)[/tex] ---- (3)
[tex](f / g)(x) = \frac{4}{5^x}[/tex] ----- (4)
[tex](f + g)(x) = -7x-7[/tex] --- (5)
[tex](f - g)(x) = 9x+ 9[/tex] --- (6)
[tex](f . g)(x) = (-8)(x+1)^2[/tex] --- (7)
[tex](f / g)(x) = -\frac{1}{8}[/tex] ----- (8)
[tex]F(t) = \frac{9}{5}t^2 + 32[/tex] ---- (9)
[tex]T(t) = (t - 2)^2 - 4[/tex] --- (10)
Step-by-step explanation:
Given
[tex]f(x) = 3[/tex]
[tex]g(x) = 4 - 6x[/tex]
Solving (1): (f + g)(x)
[tex](f + g)(x) = f(x) + g(x)[/tex]
So, we have:
[tex](f + g)(x) = 3 + 4 - 6x[/tex]
[tex](f + g)(x) = 7 - 6x[/tex]
Solving (2): (f - g)(x)
[tex](f - g)(x) =f(x) - g(x)[/tex]
So, we have:
[tex](f - g)(x) =3 - (4 - 6x)[/tex]
[tex](f - g)(x) =3 - 4 + 6x[/tex]
[tex](f - g)(x) =-1 + 6x[/tex]
[tex](f - g)(x) = 6x - 1[/tex]
Given
[tex]f(x) = 4[/tex]
[tex]g(x) = 5^x[/tex]
Solving (3): (f . g)(x)
[tex](f . g)(x) =f(x) * g(x)[/tex]
So, we have:
[tex](f . g)(x) =4 * 5^x[/tex]
[tex](f . g)(x) =4(5^x)[/tex]
Solving (4): (f / g)(x)
[tex](f / g)(x) = \frac{f(x)}{g(x)}[/tex]
So, we have:
[tex](f / g)(x) = \frac{4}{5^x}[/tex]
Given
f(x) = x + 1
g(x) = -8 - 8x
Solving (5): (f + g)(x)
[tex](f + g)(x) = f(x) + g(x)[/tex]
So, we have:
[tex](f + g)(x) = x + 1 -8-8x[/tex]
Collect Like Terms
[tex](f + g)(x) = x -8x+ 1 -8[/tex]
[tex](f + g)(x) = -7x-7[/tex]
Solving (6): (f - g)(x)
[tex](f - g)(x) = f(x) - g(x)[/tex]
So, we have:
[tex](f - g)(x) = x + 1 -( -8-8x)[/tex]
[tex](f - g)(x) = x + 1 +8+8x[/tex]
Collect Like Terms
[tex](f - g)(x) = x +8x+ 1 +8[/tex]
[tex](f - g)(x) = 9x+ 9[/tex]
Solving (7): (f . g)(x)
[tex](f . g)(x) = f(x) . g(x)[/tex]
So, we have:
[tex](f . g)(x) = (x+1) . (-8 - 8x)[/tex]
Factorize
[tex](f . g)(x) = (x+1) .(-8) (1 + x)[/tex]
Rewrite as:
[tex](f . g)(x) = (x+1) .(-8) (x+1)[/tex]
[tex](f . g)(x) = (-8)(x+1) (x+1)[/tex]
[tex](f . g)(x) = (-8)(x+1)^2[/tex]
Solving (8): (f / g)(x)
[tex](f / g)(x) = \frac{f(x)}{g(x)}[/tex]
So, we have:
[tex](f / g)(x) = \frac{(x+1)}{(-8-8x)}[/tex]
Factorize
[tex](f / g)(x) = \frac{(x+1)}{-8(1+x)}[/tex]
[tex](f / g)(x) = \frac{1}{-8}[/tex]
[tex](f / g)(x) = -\frac{1}{8}[/tex]
Solving (9):
From the question, we have that:
[tex]F(c) = \frac{9}{5}c + 32[/tex]
[tex]C(t) = t^2[/tex]
Required
Determine function F in terms of c
The implication of this question is to solve for [tex]F(c(t))[/tex]
If [tex]F(c) = \frac{9}{5}c + 32[/tex] and [tex]C(t) = t^2[/tex],
Then
[tex]F(c(t)) = \frac{9}{5}*t^2 + 32[/tex]
[tex]F(c(t)) = \frac{9}{5}t^2 + 32[/tex]
This can be rewritten as:
[tex]F(t) = \frac{9}{5}t^2 + 32[/tex]
Solving (10):
[tex]T(h) = h^2 - 4[/tex]
[tex]h(t) = t - 2[/tex]
Required
Find [tex]T(h(t))[/tex]
If [tex]T(h) = h^2 - 4[/tex] and [tex]h(t) = t - 2[/tex], then
[tex]T(h(t)) = (t - 2)^2 - 4[/tex]
This is gotten by substituting t -2 for h
The solution can be rewritten as:
[tex]T(t) = (t - 2)^2 - 4[/tex]
