Question

One more time!

If q(x)=5-x^2 and p(q(x))= 4-x^2/x^3 when x≠0, then what is p(1/4) equal to?

Can you show me the steps and possibly explain them so I understand?

Answer 1

since q(x) is inside p(x), find the x-value that results in q(x) = 1/4


`(1)/(4) = 5 - x^2\ \Rightarrow\ x^2 = 5 - (1)/(4)\ \Rightarrow\ x^2 = (19)/(4)\ \Rightarrow \\ x = (√(19) )/(2)`

so we conclude that

`q((√(19) )/(2) ) = 1/4`

therefore


`p(1/4) = p\left( q\left(( √(19) )/(2) \right) \right)`

plug
`x=√(19)/2` into p( q(x) ) to get answer


`p(1/4) = p\left( q\left( ( √(19) )/(2) \right) \right)\ \Rightarrow\ (4 - \left( (√(19) )/(2)\right)^2 )/( \left( (√(19) )/(2)\right)^3 ) \Rightarrow \\ \\ (4 - (19)/(4) )/( (19√(19) )/(8)) \Rightarrow (8\left(4 - (19)/(4)\right) )/( 8 \cdot (19√(19) )/(8)) \Rightarrow (32 - 38)/(19√(19)) \Rightarrow (-6)/(19√(19)) \cdot (√(19))/(√(19))\Rightarrow`


`(-6√(19) )/(19 \cdot 19) \\ \\ \Rightarrow -(6√(19) )/(361)`


`p(1/4) = -(6√(19) )/(361)`