Math problem master come in please?

Question

x∈R
f(x²+1)=(x²)²+5x²+2
f(x²-1)=?

Answer 1

The function f appears to be quadratic.

f(x) = ax² + bx + c

f(x² + 1) = a(x² + 1)² + b(x² + 1) + c
= ax^4 + 2ax² + 1 + bx² + b + c
= ax^4 + (2a + b)x² + (1 + b + c) = x^4 + 5x² + 2

Now we have three equations.

ax^4 = x^4
(2a + b)x² = 5x²
1 + b + c = 2

Dividing by x to the appropriate power we have:

a = 1
2a + b = 5
1 + b + c = 2

Substitute the first equation into the second.
2a + b = 5
2*1 + b = 5
b = 5 - 2 = 3

Substitute the second equation into the third.
1 + b + c = 2
1 + 3 + c = 2
c = -2

f(x) = ax² + bx + c
f(x) = x² + 3x - 2

f(x² - 1) = (x² - 1)² + 3(x² - 1) - 2
= x^4 - 2x² + 1 + 3x² - 3 - 2
= x^4 + x² - 4

Answer 2

You can rewrite the equation as
f(x^2 + 1) = (x^2 + 1)^2 + 3*(x^2 + 1) - 2

So f(x^2 - 1) = (x^2 - 1)^2 + 3*(x^2 - 1) - 2
= x^4 + x^2 - 4

Answer 3

if we put

x^2+1 as t x^2 = t-1

so f(x^2+1) = (x^2)^2 + 5x^2 + 2
or

f(t) = (t-1)^2 + 5(t-1)+ 1
= t^2 -2 t +1 + 5 t - 5 + 2 = t^2 + 3 t -2

now put t = x^2 -1 to get
f(x^2-1) = (x^2-1)^2 + 3(x^2-1) - 2

= x^4 - 2x^2 + 1 + 3x^2 - 3 -2

= x^4 +x^2 - 4

Answer 4

x^4 + x² - 4

Answer 5

f(x^2+1)=(x^2+1)^2+3(x^2+1)-2(rearranging the terms)
so f(x^2-1)=(x^2-1)^2+3(x^2-1)-2=x^4+x^2-4