Question

I am trying to figure this section out for myself before the teacher lectures on it so if some can explain this one to me would be great

Answer 1

g(f(-3))
f(x)=\[\sqrt[3]{x+1}\]
\[g(x)=4x^{2}-x\]

Answer 2

Well first sub in the value of -3 into f(x) and see what you get

Answer 3

can you first find f(-3) ?
just put x=-3 in f(x)

Answer 4

\[\sqrt[3]{-3+1}=\sqrt[3]{-2}\]
?

Answer 5

Yeah that look about right

Answer 6

Then you put that into the place of x in g(x), and solve the equation

Answer 7

so \[4\sqrt[3]{-2}^{2}-\sqrt[3]{-2}\]

Answer 8

Yeah, although that looks really messy, but it should be the right answer

Answer 9

the answer in the book says its \[4\sqrt[3]{4}+\sqrt[3]{2}\]

Answer 10

Yes they are the same form,\[\sqrt[3]{-2}^{2}=\sqrt[3]{4}\]
since you can drop the square in

Answer 11

ah yes ok get it

Answer 12

Do you get the second part also?

Answer 13

the - and - signs = + ?

Answer 14

Yes, since the minus sign is technically -1, you can bring that inside the square root and multiply it with -2, where negative and negative make positive

Answer 15

thanks :)

Answer 16

No probs :D