Inverse Functions

[asimon2005]

Hello everyone, I recently started college and I need help on these examples. I remembered that I learned them in Pre-Calculus. But some of the stuff slipped from my mind.

p.71 Sec. 1.6: #22, 26,34 a,b; 35 a

Find a Formula for the inverse of the Function
22. f(x)= 4x-1/2x+3

26. y=e^x/1+2e^x


Find the exact value of each expression.
34. (a) ln (1/e) (b) log (10) Square root(10)

35. (a) log(2)6- log (2)15+ log(2)20

I need to check to see if this one is right.

Express the given quantity as a single logarithm.

37. ln 5 + 5 ln 3

the answer I got was ln 1215. I just guess that one in my head. I used the formula lnA (UV)

 

[galactus]

37. ln 5 + 5 ln 3

the answer I got was ln 1215. I just guess that one in my head. I used the formula lnA (UV)

Remember the log law, \(\displaystyle ln(a)+ln(b)=ln(ab)\) and \(\displaystyle aln(b)=ln(b^{a})\)

\(\displaystyle ln(5)+5ln(3)=ln(5)+ln(3^{5})=ln(5\cdot 3^{3})=ln(1215)\)

You are correct. Good. It comes back to you.

BTW, please use proper grouping symbols on the other problems.

The way you have the first one written is \(\displaystyle 4x-\frac{1}{2}x+3\) when you actually mean \(\displaystyle \frac{4x-1}{2x+3}\)?.

 

[asimon2005]

Okay I got 34 and 35 but how do you do 22 and 26. It sucks that in school we are taught to analyze not memorize important material. I tell you one thing guys. I love math and science. Problem is that I use short term instead of long term. The techniques we've been taught in school is wrong. I need to find better ways on remembering this stuff in the future. If you guys have a India mantra an easy technique or something. Thanks in advance.

 

[BigGlenntheHeavy]

\(\displaystyle 22. \ f(x) \ = \ y \ = \ \frac{4x-1}{2x+3}.\)

\(\displaystyle Exchange \ the \ variables \ and \ solve \ for \ y, \ viz., \ x \ = \ \frac{4y-1}{2y+3}, \ 2xy+3x \ = \ 4y-1,\)

\(\displaystyle 2xy-4y \ = \ -3x-1, \ y \ = \ \frac{-3x-1}{2x-4} \ = \ \frac{3x+1}{4-2x} \ = \ f^{-1}(x).\)

\(\displaystyle A \ good \ check \ is \ the \ composition \ of \ f(x) \ and \ f^{-1}(x), \ as \ they \ should \ always \ reduce \ to \ x.\)

\(\displaystyle f[f^{-1}(x)] \ = \ x \ and \ f^{-1}[f(x)] \ = \ x.\)

Do the same procedure for 26.

 

[mmm4444bot]

Okay I got 34 and 35 but how do you do 22 and 26 …

… India mantra[,] an easy technique[,] or something …

There's no "easy" fix. We all gotta do the time! In other words, review. Do you still have your precalculus text? (There are also lessons on the Internet.)

Check your answers?

34(a) -1

34(b) 1/2

35(a) 3

BTW, we can type logarithms on a keyboard more clearly using one of the following conventions.

log_10[sqrt(10)] = 1/2

log_2(16) = 4

OR

log[10](sqrt[10]) = 1/2

log[2](16) = 4

Regarding inverse functions, have you ever thought about the "machine" definition of functions?

                   _______
                  |       |
     x    ---->   |   f   |    ---->    y
                  |_______|

Sometimes, it's helpful to think of "function" as a "black box" that crunches numbers; we shove a number in one end, the function changes it to something else and spits this result out the other end.

(We use the variable x to represent any number from the domain; we use the variable y to represent the resulting number in the range.)

Well, the inverse of a function reverses this process. In other words, if we have the output, but we want to know which value from the domain (input) produces it, then we need a new "black box" that will convert our y back into x.

                   _______
                  |       |
     y    ---->   |  f^-1 |    ---->    x
                  |_______|

Do you see how the input names and output names switch, when comparing these two functions f and f^-1?

So, given y = (4x - 1)/(2x + 3) as a functional relationship, the inverse is found by switching variable names and then solving for y.

x = (4y - 1)/(2y + 3)

Does this ring any bells?

After you solve for y = some expression in x, switch notation from y back to f^1(x).

Please show your work, if you need help with the algebra

Inversing an exponential involves logarithms.

In other words, solving x = e^y/(1 + 2e^y) for y will involve the natural logarithm, right? A logarithmic property is the only way to get y out of the exponent position.

Cheers

 

[asimon2005]

Gracias