Question:
The range of the real valued function \(f(x)=\dfrac{x^2+x+1}{x}\) is
The range of the real valued function \(f(x)=\dfrac{x^2+x+1}{x}\) is
Show Hint
For expressions of the form
\[
x+\frac{1}{x}
\]
remember:
\[
x+\frac{1}{x}\geq 2 \quad \text{for } x\gt 0
\]
and
\[
x+\frac{1}{x}\leq -2 \quad \text{for } x\lt 0
\]
Updated On: Jun 24, 2026
- \((-\infty,1)\cup(1,\infty)\)
- \((-\infty,-1]\cup[1,\infty)\)
- \((-\infty,-2]\cup[3,\infty)\)
- \((-\infty,-1]\cup[3,\infty)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Rewrite the given function.
Given, \[ f(x)=\frac{x^2+x+1}{x} \] Since \(x\neq 0\), \[ f(x)=x+1+\frac{1}{x} \] Let \[ y=x+1+\frac{1}{x} \]
Step 2: Separate the constant term.
We know, \[ y-1=x+\frac{1}{x} \] Now we find the range of \[ x+\frac{1}{x} \]
Step 3: Consider \(x\gt 0\).
For \(x\gt 0\), \[ x+\frac{1}{x}\geq 2 \] Therefore, \[ y-1\geq 2 \] \[ y\geq 3 \] So, for positive \(x\), \[ y\in[3,\infty) \]
Step 4: Consider \(x\lt 0\).
For \(x\lt 0\), \[ x+\frac{1}{x}\leq -2 \] Therefore, \[ y-1\leq -2 \] \[ y\leq -1 \] So, for negative \(x\), \[ y\in(-\infty,-1] \]
Step 5: Combine both ranges.
Thus, the complete range is \[ (-\infty,-1]\cup[3,\infty) \]
Step 6: Final conclusion.
Therefore, \[ \boxed{(-\infty,-1]\cup[3,\infty)} \]
Given, \[ f(x)=\frac{x^2+x+1}{x} \] Since \(x\neq 0\), \[ f(x)=x+1+\frac{1}{x} \] Let \[ y=x+1+\frac{1}{x} \]
Step 2: Separate the constant term.
We know, \[ y-1=x+\frac{1}{x} \] Now we find the range of \[ x+\frac{1}{x} \]
Step 3: Consider \(x\gt 0\).
For \(x\gt 0\), \[ x+\frac{1}{x}\geq 2 \] Therefore, \[ y-1\geq 2 \] \[ y\geq 3 \] So, for positive \(x\), \[ y\in[3,\infty) \]
Step 4: Consider \(x\lt 0\).
For \(x\lt 0\), \[ x+\frac{1}{x}\leq -2 \] Therefore, \[ y-1\leq -2 \] \[ y\leq -1 \] So, for negative \(x\), \[ y\in(-\infty,-1] \]
Step 5: Combine both ranges.
Thus, the complete range is \[ (-\infty,-1]\cup[3,\infty) \]
Step 6: Final conclusion.
Therefore, \[ \boxed{(-\infty,-1]\cup[3,\infty)} \]
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