what is the inverse of f(x)=14.95 + 3x
Hello and Welcome to OpenStudy! :) to find the inverse of a function, switch x and y so y = 14.95 + 3x becomes x = 14.95 + 3y now solve for y from there, do you know how to do that?
i did that part already, I'm stuck at what to do after that
ok, so we want y all alone on one side so first subtract 14.95 from both sides then divide both sides by 3
ok so, I'm supposed to have x-14.95/3 =y??
mmm either put a parenthesis around it like (x-14.95)/3 or distribute it so (1/3)x - (14.95)/3 = y
so thats the answer for the inverse? no more steps?
yep :)
thanks
wait i have an addition problem related to that
find f ^-1 (38.95)
so once you got f^-1(x) = (x-14.95)/3 plug in 38.95 for x
ok how do you find the range of the inverse function of f(x)=ln(x-5)+2
annnnnddddd that is something that I've deleted from my memory, sorry ummm try this http://www.wolframalpha.com/input/?i=f%28x%29%3Dln%28x-5%29%2B2+range but I won't be able to explain it for you, sorry
how bout the y intercept of f(x)=ln(x-5)+2 , its fine if you don't know
ahhhh mk that's something I kinda know the parent function is y = ln(x) how it is moved around is y = ln(x-h) + k where h is how much left <-> right and k is how much up down orrrr you can think y = mx + b and b = y-intercept same idea for this format... so 2 is the y-intercept
i have to show my work , so can you explain a little more how you arrived at 2
oh... show work... thennnn y-intercept is when x = 0 plug in 0 for x and simplify
k thanks
*scratches head* I guess it's not 2 then... *shrugs* I failed algebra 2 xD but the thing about y-intercept is when x=0 should be right
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