Suppose f(t^2+1)=t^4+5t^2+3. What is f(t^2-1)?
f(t^2+1) = t^4+5t^2+3 = (t^2+1)(t^2+4) - 1
= (t^2+1)((t^2+1)+3)-1
That means that f(t) = t(t+3)-1
So,
f(t^2-1) = (t^2-1)((t^2-1)+3)-1 = t^4+t^2-3
Thanks Steve, Could you please explain more on these two last steps:
That means that f(t) = t(t+3)-1
So,
f(t^2-1) = (t^2-1)((t^2-1)+3)-1 = t^4+t^2-3
To find the value of f(t^2-1) using the given equation f(t^2+1)=t^4+5t^2+3, we need to substitute t^2-1 in place of t^2 in the equation.
Let's replace t^2 with (t^2-1) in the given equation:
f((t^2-1)+1)=(t^2-1)^4+5(t^2-1)^2+3
Simplifying the equation, we get:
f(t^2)=t^8 - 4t^6 + 6t^4 - 4t^2 + 9
Therefore, substituting t^2-1 for t^2, we find:
f(t^2-1)=(t^2-1)^8 - 4(t^2-1)^6 + 6(t^2-1)^4 - 4(t^2-1)^2 + 9
Simplifying this equation will give us the value of f(t^2-1).