ｔｈｅｒｅ　ａｒｅ　ｔｗｏ　ｆｕｎｃｔｉｏｎｓ：
ｆ（ｘ）＝（１／８）ｘ－３　　　　ａｎｄ　　　ｇ（ｘ）＝ｘ＾３

ｈｏｗ　ｄｏ　ｙｏｕ　ｆｉｎｄ　ｆ＾－１（ｇ＾－１（１））？

2. Originally Posted by got_jane
ｔｈｅｒｅ　ａｒｅ　ｔｗｏ　ｆｕｎｃｔｉｏｎｓ：
ｆ（ｘ）＝（１／８）ｘ－３　　　　ａｎｄ　　　ｇ（ｘ）＝ｘ＾３

ｈｏｗ　ｄｏ　ｙｏｕ　ｆｉｎｄ　ｆ＾－１（ｇ＾－１（１））？
why don't you type with spaces?

to find $f^{-1}(x)$, start with
$y = \frac 18x - 3$

switch x and y and solve for y. that will give you $f^{-1}(x)$. do a similar thing to find $g^{-1}(x)$.

can you continue?

3. Originally Posted by Jhevon
why don't you type with spaces?

to find $f^{-1}(x)$, start with
$y = \frac 18x - 3$

switch x and y and solve for y. that will give you $f^{-1}(x)$. do a similar thing to find $g^{-1}(x)$.

can you continue?
$f^{-1}(x)=8x+3$ and $g^{-1}(x)=x^{1/3}$
and so $f^{-1}(g^{-1}(1))=8(1)^{1/3}+3$, I get 11 as an answer, but I checked the back of the book the correct answer is supposed to be 32...I wonder what I'm doing wrong.

4. Originally Posted by got_jane
$f^{-1}(x)=8x+3$ and $g^{-1}(x)=x^{1/3}$
and so $f^{-1}(g^{-1}(1))=8(1)^{1/3}+3$, I get 11 as an answer, but I checked the back of the book the correct answer is supposed to be 32...I wonder what I'm doing wrong.
your function for $f^{-1}(x)$ is wrong. it is $8(x + 3) = 8x + 24$

5. Originally Posted by Jhevon
your function for $f^{-1}(x)$ is wrong. it is $8(x + 3) = 8x + 24$
oh! I got it! thank you so much! i appreciate it!