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Inverse Functions Question

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I do not fully understand how to complete these types of problems:

If $$f(x) = \frac {x^5 - 1} {3},$$ find $$f^-1 (-31 / 96)$$

Feb 21, 2018

#1
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First , let's find the inverse function f^-1

change x's to y's and y's (which is f(x) )    to x's       then solve for y

x= (y^5-1) / 3

(3x+1)^1/5 = y      Now substitue  -31/96   in to the equation

(3 (-31/96) +1 )^1/5  = y   = 0.5

Feb 21, 2018
#2
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A question for ElectricPavlov

What would your answer have been if this person wanted   $$F^{-1}(-11)\;\;\;$$

Melody  Feb 21, 2018
#3
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Well....with what I have done so far   f^-1 (-11)  = -2

Did I get something incredibly incorrect?  Am I going to be suspended? Ridiculed? Embarrassed?  Imprisoned?  Caned?  Beheaded?  ?

Not sure why you are asking?  Hmmmmmm.......

ElectricPavlov  Feb 21, 2018
#4
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Hi G

I remember when I first had problems with inverse functions and there are many out there that still do so dont feel too bad as it is be a difficult topic to master. Hope it helps, what I did was firstly read over the definition a number of times until it made sense to me then I went over some really simple examples and drew graphs of each situation to further help things sink in...basically like many mathematical concepts you need to devote a lot of time and effort and oftem frustration until it sinks in and then when finally the penny drops and it will you will wonder why it took so long because it now seems so easy. One piece of information which helped is to know that inverse functions as graphs are reflections of one another in the line y=x as you can see in the graph given by EP. Lets look at a reasonably simple example. Find the inverse function of y=x2. Oh by the way another gem...it needs to be a function as the name implies....Really what all the fuss is about who knows as all you are doing is swapping the x values with the y values for each point in question ie

if y = x2  then the inverse is x = y2 and so we use algebra to rearrange this to make y the subject ie y=sqr(x) or

y=-sqr(x). Now if you sketch these two functions you get your standard parabola and the same parabola rotated through 90° to the right. Now draw the line y=x. you will see that y=x2 and x=y2 are reflections in y=x. However, x=y2is not a function but y=sqr(x) ie positive root is. So the inverse function of y=xis by definition the positive root y=sqr(x) and the fact that y=x2 has a range of y>=0 which becomes the domain of the inverse ie y=sqr(x),x>=0.

Also look for points of intersection to help find the inverrse function.

Hope this was a bit of a help. Remember the key is practice practice practice!!!

Feb 21, 2018
edited by Guest  Feb 21, 2018
#5
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Hi EP,

It might hlp the guest that asked though.

What I was getting at is this  (but it makes no differnence for this question)

The definition of a function is that for every x vlaue there can be at most 1 y value.

If it is graphed this is easy to see from the straight line test.

The inverse of a funtion is its reflection in the line y=x

When you do this, most times the reflection is NOT a function, so you need to be very careful about the new domain.

All the teachers I have ever discussed this with say you must rearrange the equation to make x the subject BEFORE you swap the x with the y BECAUSE then the new domain and range are more easily seen.

I cannot see why they make this demand so strongy but certainly people must be extremely careful about constraints.

say if

y=x^2  this is a normal concave up parabola

If you just swap the lpronumerals like you said then you get   x=y^2    This is a sideways parabola BUT it is not the inverse function of y=x^2 because it is NOT a function at all.

To find the inverse of y=x^2 you should rearange the formula first and make sure that for each y value there is ONLY one x.

THEN you can swap the letters over.

$$y=x^2\\ \pm\sqrt{y}=x\\ x=\pm\sqrt{y}\\ \text{It can easily be seen that } y\ge0\\ \text{Swap pronumerals and drop the minus sign}\\ y=\sqrt x \qquad \text{If you include both answers this will not be a function }\\ \text{The inverse of }y=x^2 \text{ is } y=\sqrt{x} \quad x\ge 0 ,\quad y\ge0$$

Here is the graph of y=x^2 in red and its inverse function in blue

Feb 22, 2018
#6
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Cool!  I understand...thanx for the clarification, Melody.....I'll remember that for future Q and As !!

Feb 22, 2018
#7
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OH Darn! I was hoping to see an execution or a slaughter. There’s nothing like a math fight where the only thing remaining of the gladiators are fractions and prime numbers.  Oh well, maybe another day.

Feb 22, 2018
#8
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Sorry to disappoint you

Melody  Feb 23, 2018
#9
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Well maybe next time then. I remember a few times you got into it with Nauseated. These were always entertaining...and sometimes gory.  I miss the good-old-days.

Guest Feb 23, 2018
#10
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Yea me too. Although some of the gore I could have done without LOL.

Melody  Feb 23, 2018