Answer:
(-2,-2)
Step-by-step explanation:
It’s B I hope this helps :)
Answer:
![f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx-4%7D%20%2C%20g%28x%29%3D6x%5E%7B2%7D%5Ctextrm%7B%20or%20%7Df%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%2Cg%28x%29%3D6x%5E%7B2%7D%20-4)
Step-by-step explanation:
Given:
The function, ![H(x)=\sqrt[3]{6x^{2}-4}](https://tex.z-dn.net/?f=H%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D)
Solution 1:
Let ![f(x)=\sqrt[3]{x}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D)
If ![f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3DH%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D)
, then,
![\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bg%28x%29%7D%20%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5Cg%28x%29%3D6x%5E%7B2%7D-4)
Solution 2:
Let ![f(x)=\sqrt[3]{x-4}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx-4%7D)
. Then,
![f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3DH%28x%29%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%5C%5C%5Csqrt%5B3%5D%7Bg%28x%29-4%7D%3D%5Csqrt%5B3%5D%7B6x%5E%7B2%7D-4%7D%20%5C%5Cg%28x%29-4%3D6x%5E%7B2%7D-4%5C%5Cg%28x%29%3D6x%5E%7B2%7D)
Similarly, there can be many solutions.
Answer:
To do this, you need to multiply out the expressions. This is a bit tedious, but remember like FOIL for binomials, for these trinomials you must multiply each term. If you need a step-by-step, I'd be happy to provide it. Let me know.
Once you have simplified the expression, you get
-x-9/2x-4
But, the problem stipulates that a must equal 1. We can equivalently factor out the negative sign and put it on the denominator with no change to write
x+9/-(2x-4) = x+9/-2x+4
So, seeing where each coefficient corresponds between the two expressions, you get a = 1, b = 9, c = –2, and d = 4.
