Question

Can anyone answer me this question ​

Answer 1

Question 1:

For this case we have the following functions:

$f (x) = 4x + 1\\g (x) = x ^ 3 + 1$

We must find$g_ {o} f (0$):

By definition we have to:

$g_ {o} f = g (f (x))\\f_ {o} g = f (g (x))$

$g (f (x)) = (4x + 1) ^ 3 + 1$

We substitute $x = 0$:

$g (f (0)) = (4 (0) +1) ^ 3 + 1 = 1 ^ 3 + 1 = 2$

So, we have that $g (f (0)) = 2$

Answer:

$g (f (0)) = 2$

Question 2:

For this case we have the following functions:

$f (x) = 4x + 1\\g (x) = x ^ 3 + 1$

We must find $f_ {o} g (0)$:

By definition we have to:

$f_ {o} g = f (g (x))\\f (g (x)) = 4 (x ^ 3 + 1) + 1 = 4x ^ 3 + 4 + 1 = 4x ^ 3 + 5$

We substitute $x = 0$:

$f (g (0)) = 4 (0) ^ 3 + 5 = 5$

Answer:

$f (g (0)) = 5$

Question 3:

For this case we must find the inverse of the following function:

$h (x) = \frac {2x + 1} {3}$

To do this, we follow the steps below:

We change y for $h (x)$:

$y = \frac {2x + 1} {3}$

We exchange variables:

$x = \frac {2y + 1} {3}$

We clear the value of the variable "y":

$3x = 2y + 1\\3x-1 = 2y\\y = \frac {3x} {2} - \frac {1} {2}$

We change y for $h^{ -1} (x)$:

$h ^{ - 1} (x) = \frac {3x} {2} - \frac {1} {2}$

Answer:

$h ^ {- 1} (x) = \frac {3x} {2} - \frac {1} {2}$