Numerical answer
A function f, defined for all non-zero real numbers, satisfies f(x) + 2f(1/x) = 3x for every x ≠ 0. What is the sum of all real values of x for which f(x) = 1?
Show answer and solution
Answer: -1
Replace x by 1/x: f(1/x) + 2f(x) = 3/x.
Now we have two equations in f(x) and f(1/x):
(i) f(x) + 2f(1/x) = 3x
(ii) 2f(x) + f(1/x) = 3/x
2 × (ii) − (i): 3f(x) = 6/x − 3x ⇒ f(x) = 2/x − x.
Set f(x) = 1: 2/x − x = 1 ⇒ 2 − x² = x ⇒ x² + x − 2 = 0 ⇒ (x + 2)(x − 1) = 0 ⇒ x = 1 or x = −2.
Both are non-zero, so both are valid. Sum = 1 + (−2) = −1.
Tip: For functional equations involving f(x) and f(1/x), substitute x → 1/x and solve the resulting pair of linear equations.
Now we have two equations in f(x) and f(1/x):
(i) f(x) + 2f(1/x) = 3x
(ii) 2f(x) + f(1/x) = 3/x
2 × (ii) − (i): 3f(x) = 6/x − 3x ⇒ f(x) = 2/x − x.
Set f(x) = 1: 2/x − x = 1 ⇒ 2 − x² = x ⇒ x² + x − 2 = 0 ⇒ (x + 2)(x − 1) = 0 ⇒ x = 1 or x = −2.
Both are non-zero, so both are valid. Sum = 1 + (−2) = −1.
Tip: For functional equations involving f(x) and f(1/x), substitute x → 1/x and solve the resulting pair of linear equations.
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