Algebra · real student question

Simplify f(x) = (-x2 - 289)/x and state its domain.

Question

Simplify f(x)=x2289xf(x)=\dfrac{-x^2-289}{x} and state the domain.

Step-by-step solution

  1. Check whether the numerator factors first. x2289=(x2+289)-x^2-289 = -(x^2+289). Since x2+289x^2+289 is a sum of squares it has no real factors, so nothing cancels wholesale - splitting the fraction is the only simplification available.

  2. Split the single fraction into two. Division distributes over the terms of the numerator: x2289x=x2x289x.\frac{-x^2-289}{x} = \frac{-x^2}{x} - \frac{289}{x}. This is legal because both pieces keep the same denominator xx.

  3. Cancel one factor of x in the first term. x2x=xxx=x\dfrac{-x^2}{x} = \dfrac{-x\cdot x}{x} = -x, valid because we are already assuming x0x \neq 0. The second term has no xx in its numerator, so it stays as 289x-\dfrac{289}{x}.

  4. Assemble the simplified form. f(x)=x289x=(x+289x).f(x) = -x - \frac{289}{x} = -\left(x + \frac{289}{x}\right). The bracketed form makes the structure obvious: ff is 1-1 times a number-plus-its-reciprocal-multiple.

  5. State the domain. The original denominator is xx, so x=0x=0 must be excluded; the domain is x(,0)(0,)x \in (-\infty,0)\cup(0,\infty). Note that the simplified form has the same restriction, so no hole was created or destroyed.

  6. Read off the structure as a bonus. Since f(x)=x+289x=f(x)f(-x) = x + \dfrac{289}{x} = -f(x), the function is odd, and x=0x = 0 is a vertical asymptote while y=xy = -x is a slant asymptote.

Answer

f(x)=x289x=(x+289x),x0f(x) = -x - \frac{289}{x} = -\left(x+\frac{289}{x}\right), \qquad x \neq 0

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