Let f(x) = x2 - 16. Find f-^1(x). Please, someone help. i don't need the answer, just help.
does it state that x > 0 or x < 0 or something like that?
no, everything that I see for the problem is what I have above. :/
ok usually they'll add the restriction that x > 0, so let's go with that
f(x) = x^2 - 16 y = x^2 - 16 ... replace f(x) with y x = y^2 - 16 ... swap x and y now solve for y and tell me what you get
I'd hate to sound stupid, but would it be along the lines of something like 16x = y^2?
you don't sound stupid
first you need to move that 16 over, so you add 16 to both sides x = y^2 - 16 x + 16 = y^2
oohhhh okay. Instead I just made it 16x.
my bad. :P
so y^2 = x + 16 then you take the square root of both sides because x > 0 is the domain of f(x), this means that the range of f^-1(x) is y > 0 (remember domain and range swap when you go from a function to its inverse) so because y > 0, this means that we go with the positive square root, so y = sqrt(x+16)
therefore, if x > 0 is applied to the original function, then the inverse is \[\large f^{-1}(x) = \sqrt{x+16}\]
That's one of the answers! Thanks so much! But for future reference, what would I have to remember in order to solve equations like these? I just started working with this kind of stuff, and I don't exactly understand it too well.
when solving equations you are following PEMDAS in reverse so normally if you started with something like 2*3 + 5, you follow PEMDAS in the forward direction to get 2*3+5 6+5 11 so 2*3+5 = 11 now say that the '3' is covered up and turned into an 'x' to get 2x+5 = 11 ----------------- now you follow PEMDAS in reverse to undo all the operations to get x all by itself 2x + 5 = 11 2x = 11 - 5 ... (undoing addition) 2x = 6 x = 6/2 .... (undoing multiplication) x = 3
notice how multiplication normally comes first in PEMDAS but because we're following it in reverse order, addition comes first (ie that's the first thing to undo)
so looking at x = y^2 - 16 the first thing to undo is the subtraction of 16, so we add 16 to both sides to get x+16 = y^2 y^2 = x+16 then we undo the exponent by taking the square root of both sides y = sqrt(x+16)
normally exponentiation is done before subtraction (since the E comes before S), but again we're going in reverse when solving for a variable
ohhh, so I have to find the square root, only if there are exponents. I was so confused on when it was necessary to find the square root. thanks, that clears things up a little more for me. (:
only if the exponent is 2 ex: x^2 = 4 x = sqrt(4) or x = -sqrt(4) x = 2 or x = -2 but if the exponent is 3 like in x^3 = 8, then you use a cube root to solve for x
yeah, that makes sense. I just started so I've only seen exponents of 2 so far, but that's good to know for future reference.(:
ok glad it's making more sense
If you're not busy, can I have your help with one more question?
sure go for it
Everything you've explained to me, makes complete sense. But... Let f(x) = x - 2 and g(x) = x2 - 7x - 9. Find f(g(-1)).
first we need to find g(-1)
what is g(-1)
I have no idea, I'm sorry. :( would i just solve the equation g(x) = x2 - 7x - 9?
g(-1) means plug x = -1 into g(x)
g(x) = x^2 - 7x - 9 g(-1) = (-1)^2 - 7(-1) - 9 g(-1) = ??
g(-1) = 1+7-9 am i on the right track?
yep, keep going
Then g(-1)+9=8, right?
why not just combine 1+7-9 to get ______
ohhh, hahaha that's obvious, sorry. -1 :)
so g(-1) = -1, good
so f(g(-1)) turns into f(-1) since you replace g(-1) with -1 (we've just shown that g(-1) = -1)
now you must find f(-1)
f(x) = x - 2 f(-1) = (-1) - 2 f(-1) = -3
very good, that means f(g(-1)) = -3
oh wow!! that's it?
yep, that's all there is to it
hahaha i guess it just looked over whelming. thank you so much again. :) i really really appreciate it.
you're welcome, I'm glad I could help
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