givin the fuction if f(x)=3x-4/5 which of the below expressions are correct? f−1(x) = The quantity of 5x plus 4, divided by 3. f−1(x) = The quantity of 5x minus 4, divided by 3. f−1(x) = The quantity of negative 3x minus 4, divided by 5. f−1(x) = The quantity of 4 minus 3x, divided by 5.
@johnweldon1993
Okay so this in inverse functions is this \[\large f(x) = 3x - \frac{4}{5}\] or \[\large f(x) = \frac{3x - 4}{5}\]
okayy
There was a question in that :P
ohh my bad lol its the bottom one sorry
Lol alright so we have \[\large y = \frac{3x - 4}{5}\] Im sure you know that f(x) can just be written as 'y' right?
yes sir :)
Good..so yes we have \[\large y = \frac{3x - 4}{5}\] what the basic idea of inverse function is...you switch the 'x' ad the 'y'....then you solve again for 'y' so...lets switch x and y \[\large x = \frac{3y - 4}{5}\]
And now we just need to use algebra to solve this for 'y' again...how would we do that?
multiply the x by 5 nd 3y-4 by 5?
Correct....leaving us with \[\large 5x = 3y - 4\] correct?
yeppers
What next? we need 'y' all by itself
divide all of the numbers by 3 or is it just 5/3
Hmm...well yeah we could do that first....but maybe we should add 4 to both sides first ...what do ya think?
suree which ever way is easier for me lol
you would get 8x = 3y right ?
Well we COULD do it either way...lol but now you've forced me to show you both ways :P
nahh just do anyone idc and i didnt force you :p
Lol :P alright well we'll do it the other way I said \[\large 5x = 3y - 4\] after you add 4 to both sides you get \[\large 5x + 4 = 3y\] right?
yaa
And finally you would divide everything by 3 to solve for 'y' \[\large f^{-1}(x) = \frac{5x + 4}{3}\]
Omg i love you forever jk u are amazing
Lol <3 haha glad to help :)
can i tag u if i have anymore questions please u make me feel not stupid
Of course :)
thankss
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