Question

01.02 Introduction to Functions... I need help

Answer 1

Answer:

$f {}^{ - 1} (x)$

means the inverse function of

$f(x)$

To do that, we just do the following steps.

1. Set up equation

$f(x) = 2x + 2$

Remember Ruler Notation, that f(x)=y

$y = 2x + 2$

Swap x and y

$x = 2y + 2$

Solve for y

$x - 2 = 2y$

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

So we found our inverse function, now we just plug in 4 for x

$( \frac{1}{2} )(4) - 1 = y$

$2 - 1 = y$

$1 = y$

Option 4 is the answer.

For question 6, plug in 2 for x in function g

We get

$(2) {}^{2} - 6(2) - 7$

$= - 15$

Now, we plug in -15 for f.

$- 15 + 8$

$= - 7$

Option 1 is the answer.

For question 7, plug in g(x) for x in f(x).

$9( - x + 3) - 2 = - 9x + 27 - 2 = - 9x + 25$

Option 1 is the answer.

We are given

${x}^{2} - 16 = y$

${y}^{2} - 16 = x$

$y {}^{2} = x + 16$

$y = \sqrt{x + 16}$

Option 4 is the answer. Rember square roots give two answers, a positive and negative unless we take the only the positive answer.

Plug in -1 in g.

$( - 1) {}^{2} - (7)( - 1) - 9 = - 1$

Now, plug that in for x in f(x).

$- 1 - 2 = - 3$

Option 1 is the answer.

Next, just compose again.

$- 4(10x - 6) + 7 = - 40x + 24 + 7 = - 40x + 31$

Option 4 is the answer.

$y = 2x - 6$

$y + 6 = 2x$

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

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

Plug in 2

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

$y = 4$

First Option is answer.