Algebra · real student question

Express the domain of f(x) = (2x + 1)/(x^4 - 5x^2 + 4) in interval notation.

Question

Express the domain of

f(x)=2x+1x45x2+4f(x) = \frac{2x+1}{x^4 - 5x^2 + 4}

in interval notation.

Step-by-step solution

  1. Identify the only restriction. The numerator is a polynomial, defined everywhere; the function fails only where the denominator is zero. So the task reduces to solving x45x2+4=0x^4 - 5x^2 + 4 = 0.

  2. Solve the biquadratic with a substitution. Only even powers appear, so put u=x2u = x^2:

    u25u+4=0(u1)(u4)=0u=1 or u=4u^2 - 5u + 4 = 0 \quad\Longrightarrow\quad (u-1)(u-4) = 0 \quad\Longrightarrow\quad u = 1 \text{ or } u = 4

  3. Return to x. Each positive uu gives two real values of xx:

    x2=1x=±1,x2=4x=±2x^2 = 1 \Rightarrow x = \pm1, \qquad x^2 = 4 \Rightarrow x = \pm2

    Four excluded points in total — the maximum a quartic denominator can produce.

  4. Remove the four points and write the intervals. Ordered on the number line, the excluded values are 2,1,1,2-2, -1, 1, 2, which cut R\mathbb{R} into five pieces:

    (,2)(2,1)(1,1)(1,2)(2,)(-\infty,-2) \cup (-2,-1) \cup (-1,1) \cup (1,2) \cup (2,\infty)

  5. Check one point in each interval. At x=3x = -3: denominator 8145+4=40081 - 45 + 4 = 40 \ne 0. At x=1.5x = -1.5: 5.062511.25+4=2.18755.0625 - 11.25 + 4 = -2.1875. At x=0x = 0: 44. At x=1.5x = 1.5: 2.1875-2.1875. At x=3x = 3: 4040. None vanishes, confirming the five intervals are entirely inside the domain.

Answer

(,2)(2,1)(1,1)(1,2)(2,)(-\infty,-2) \cup (-2,-1) \cup (-1,1) \cup (1,2) \cup (2,\infty)

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