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Zorluk: ZorFunctions and Custom Symbol Operations

For all non-zero real numbers xx, the function ff satisfies the equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x. What is the value of f(2)f(2)?

  1. −1-1Cevap
  2. B
    00
  3. C
    11
  4. D
    22
  5. E
    33

Cevap

−1-1
Evaluating the functional equation at x=2x = 2 gives f(2)+2f(1/2)=6f(2) + 2f(1/2) = 6. Evaluating the equation at x=1/2x = 1/2 gives f(1/2)+2f(2)=3/2f(1/2) + 2f(2) = 3/2. Substituting f(1/2)=3/2−2f(2)f(1/2) = 3/2 - 2f(2) into the first equation produces −3f(2)+3=6-3f(2) + 3 = 6, which simplifies to f(2)=−1f(2) = -1.

Adım Adım Çözüm

1
Substitute x=2x = 2 into the given functional equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x.
f(2)+2f(12)=6f(2) + 2f\left(\frac{1}{2}\right) = 6
This establishes a relation between the target value f(2)f(2) and the auxiliary value f(12)f\left(\frac{1}{2}\right).
2
Substitute x=12x = \frac{1}{2} into the original equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x.
f(12)+2f(2)=3(12)=32f\left(\frac{1}{2}\right) + 2f(2) = 3\left(\frac{1}{2}\right) = \frac{3}{2}
This provides a second independent linear equation relating f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
3
Isolate f(12)f\left(\frac{1}{2}\right) in the second equation: f(12)=32−2f(2)f\left(\frac{1}{2}\right) = \frac{3}{2} - 2f(2), and substitute this expression into the first equation.
f(2)+2(32−2f(2))=6  ⟹  f(2)+3−4f(2)=6  ⟹  −3f(2)+3=6f(2) + 2\left(\frac{3}{2} - 2f(2)\right) = 6 \implies f(2) + 3 - 4f(2) = 6 \implies -3f(2) + 3 = 6
Eliminating f(12)f\left(\frac{1}{2}\right) converts the system into a single linear equation in terms of f(2)f(2).
4
Solve −3f(2)+3=6-3f(2) + 3 = 6 for f(2)f(2).
−3f(2)=3  ⟹  f(2)=−1-3f(2) = 3 \implies f(2) = -1
Subtracting 33 from both sides gives −3f(2)=3-3f(2) = 3, and dividing by −3-3 yields f(2)=−1f(2) = -1.

Anahtar Kavram

Solving Functional Equations via Variable Substitution

Alternatif Yöntem

Generalize f(x)f(x) by setting f(x)=Ax+Bxf(x) = Ax + \frac{B}{x}. Substituting into f(x)+2f(1/x)=3xf(x) + 2f(1/x) = 3x gives (Ax+B/x)+2(A/x+Bx)=(A+2B)x+(2A+B)/x=3x(Ax + B/x) + 2(A/x + Bx) = (A + 2B)x + (2A + B)/x = 3x. Equating coefficients: A+2B=3A + 2B = 3 and 2A+B=0  ⟹  B=−2A  ⟹  A−4A=3  ⟹  A=−1,B=22A + B = 0 \implies B = -2A \implies A - 4A = 3 \implies A = -1, B = 2. Thus f(x)=−x+2/xf(x) = -x + 2/x, so f(2)=−2+2/2=−1f(2) = -2 + 2/2 = -1.
Tahmini Süre:1m 30s
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Functions and Custom Symbol Operations

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