1) Define g by g(x) = 5x - 4. If g(x) = f^-1(x) - 3 and f^-1(x) is the inverse of the function f(x) = ax+b, find 5a + 5b.

So far, I've found the inverse of f, which is (x-b)/a. The equation would be $5x - 4 = (x-b)/a - 3$. Please help me find $a$ and $b$ to get $5a + 5b$. Thank you so much!

(I completely do not get the second question...I don't understand what the brackets mean)

2) Let g(x) be a function piecewise defined as $f(x) = \left\{ \begin{array}{c l} -x & x\le 0 \\ 2x-41 & x>0 \end{array} \right..$

If $a$ is negative, find $a$ so that $g(g(g(10.5))) = g(g(g(a)))$. Express your answer as a decimal.

(If you can only answer one of the two questions, that is fine!)

Thank you so much!

PartialMathematician Nov 18, 2018

#1**+3 **

f(x) = ax + b for f(x), write y

y = ax + b subtract b rom both sides

[y - b] = ax divide both sides by a

[ y - b] / a = x "swap' x and y

[x - b] / a = y

(1/a)x - b/a = y = f^{-1}(x)

So g(x) = f^{-1}(x) - 3 = (1/a)x - b/a - 3 = 5x - 4

This implies that

(1/a) = 5 and -b/a - 3 = 4

So

a = (1/5) and

-b / (1/5) - 3 = -4 add 3 to both sides

-b / (1/5) = -1

-b = -1 (1/5)

-b = -1/5

b = 1/5

So

5a + 5b =

5(1/5) + 5(1/5) = 1 + 1 = 2

CPhill Nov 18, 2018

#2**0 **

Thanks so much, CPhill! I fully understand the solution now!

Solution to second question is \(a = -30.5\).

PartialMathematician
Nov 18, 2018

#3**+2 **

**Hi PartialMathematician, welcome to the forum.**

In a couple of years time are you going to change your name to FullMathematician ? LOL

On a serious note:

Please put some effort into learning to presenting your questions properly.

The stuff between the dollar signs is LaTex code.

If you delete the dollar signs you can past the rest into the LaTex box that will appear when you click [LaTex] in the ribbon.

After you hit ok it will present properly.

Note:

Just delete the stuff that is in the box when you open it.

Melody Nov 18, 2018