Suppose f is one-to-one, f(7) = 8, and f'(7) = 8. Find a) f^-1(3) and b) f^-1'(3). I'm kinda stuck on how to find part b). Also check my work for part a) ?
does this problem come with a graph?
y-8 = 8(x-7) y = 8x+48 so f^-1(x) = x = 8y - 48 so x = 8y - 48 so 8y = x+48 so y = (x+48)/8 so f^-1(3) = 51/8 = 6.375
no.
ok you found \[y=\frac{ x+48 }{ 8 }\] for the inverse function right? do you know how to take the derivatives yet?
yeah. -_- OMG. it's that easy... my teacher showed us a TOTALLY confusing way to do it. so i guess i just got too confused. so i think i just answered my own question. how sad. :(
haha it's ok :) glad you figured out how to do it!
is there a way to do this without finding the inverse function?
yes
would you just take the derivative of the original function and flip it? and that would be the derivative of the inverse? idk just a guess.
but i have a question how do you know the function is a line?
it doesn't say anything about it being a line if it was a line, they would say \(f'(x)=8\) not \(f'(7)=8\)
lol i don't. i just assumed. ohhhhh now i see it.
in fact there must be some other part missing in the question knowing that \(f(8)=7\) and that \(f'(8)=7\) tells you nothing about this function at 3 or its inverse at 3
it just says find inverse and inverse derivative at 3. that's all that's given; what i typed.
can you take a picture of the problem?
lol my teacher got it from a random website and she posted the problem on her website. and i just copy and pasted whatever she had. THERE IS NO GRAPH. Suppose f is one-to-one, \[f(7) = 8\] \[f'(7) = 8\] Find a) \[f^{-1}(3) \] b) \[f'^{-1}(3)\] I just wrote it in a better format. i took a picture anyways. it's uploading.
what if she was just trolling you when the real answer would be "cannot conclude answer without enough information"
ignore my work and other stuff stuff. it's #2.
i feel like she copied the problem wrong. i don't know any way to acquire this without assuming the equation was linear :( sorry
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