Question

Can you help me I have no idea how to do this

Answer 1

Answer:

$(f + g)(x) = 7 - 6x$ --- (1)

$(f - g)(x) = 6x - 1$ -- (2)

$(f . g)(x) =4(5^x)$ ---- (3)

$(f / g)(x) = \frac{4}{5^x}$ ----- (4)

$(f + g)(x) = -7x-7$ --- (5)

$(f - g)(x) = 9x+ 9$ --- (6)

$(f . g)(x) = (-8)(x+1)^2$ --- (7)

$(f / g)(x) = -\frac{1}{8}$ ----- (8)

$F(t) = \frac{9}{5}t^2 + 32$ ---- (9)

$T(t) = (t - 2)^2 - 4$ --- (10)

Step-by-step explanation:

Given

$f(x) = 3$

$g(x) = 4 - 6x$

Solving (1): (f + g)(x)

$(f + g)(x) = f(x) + g(x)$

So, we have:

$(f + g)(x) = 3 + 4 - 6x$

$(f + g)(x) = 7 - 6x$

Solving (2): (f - g)(x)

$(f - g)(x) =f(x) - g(x)$

So, we have:

$(f - g)(x) =3 - (4 - 6x)$

$(f - g)(x) =3 - 4 + 6x$

$(f - g)(x) =-1 + 6x$

$(f - g)(x) = 6x - 1$

Given

$f(x) = 4$

$g(x) = 5^x$

Solving (3): (f . g)(x)

$(f . g)(x) =f(x) * g(x)$

So, we have:

$(f . g)(x) =4 * 5^x$

$(f . g)(x) =4(5^x)$

Solving (4): (f / g)(x)

$(f / g)(x) = \frac{f(x)}{g(x)}$

So, we have:

$(f / g)(x) = \frac{4}{5^x}$

Given

f(x) = x + 1

g(x) = -8 - 8x

Solving (5): (f + g)(x)

$(f + g)(x) = f(x) + g(x)$

So, we have:

$(f + g)(x) = x + 1 -8-8x$

Collect Like Terms

$(f + g)(x) = x -8x+ 1 -8$

$(f + g)(x) = -7x-7$

Solving (6): (f - g)(x)

$(f - g)(x) = f(x) - g(x)$

So, we have:

$(f - g)(x) = x + 1 -( -8-8x)$

$(f - g)(x) = x + 1 +8+8x$

Collect Like Terms

$(f - g)(x) = x +8x+ 1 +8$

$(f - g)(x) = 9x+ 9$

Solving (7): (f . g)(x)

$(f . g)(x) = f(x) . g(x)$

So, we have:

$(f . g)(x) = (x+1) . (-8 - 8x)$

Factorize

$(f . g)(x) = (x+1) .(-8) (1 + x)$

Rewrite as:

$(f . g)(x) = (x+1) .(-8) (x+1)$

$(f . g)(x) = (-8)(x+1) (x+1)$

$(f . g)(x) = (-8)(x+1)^2$

Solving (8): (f / g)(x)

$(f / g)(x) = \frac{f(x)}{g(x)}$

So, we have:

$(f / g)(x) = \frac{(x+1)}{(-8-8x)}$

Factorize

$(f / g)(x) = \frac{(x+1)}{-8(1+x)}$

$(f / g)(x) = \frac{1}{-8}$

$(f / g)(x) = -\frac{1}{8}$

Solving (9):

From the question, we have that:

$F(c) = \frac{9}{5}c + 32$

$C(t) = t^2$

Required

Determine function F in terms of c

The implication of this question is to solve for $F(c(t))$

If $F(c) = \frac{9}{5}c + 32$ and $C(t) = t^2$,

Then

$F(c(t)) = \frac{9}{5}*t^2 + 32$

$F(c(t)) = \frac{9}{5}t^2 + 32$

This can be rewritten as:

$F(t) = \frac{9}{5}t^2 + 32$

Solving (10):

$T(h) = h^2 - 4$

$h(t) = t - 2$

Required

Find $T(h(t))$

If $T(h) = h^2 - 4$  and   $h(t) = t - 2$, then

$T(h(t)) = (t - 2)^2 - 4$

This is gotten by substituting t -2 for h

The solution can be rewritten as:

$T(t) = (t - 2)^2 - 4$