Question

The range of the function \[ f(x)=\frac{x}{x^2-5x+9} \] is:

• A. \[ \left[\frac{1}{11},1\right] \]
• B. \[ \left[-\frac{1}{11},1\right] \]
• C. \[ \left[-1,-\frac{1}{11}\right] \]
• D. \[ \left[-1,\frac{1}{11}\right] \]

Hint:

To find the range of a rational function, put \(y=f(x)\), rearrange into a quadratic in \(x\), and use the condition \(\Delta \geq 0\) for real solutions.

Answer 1

The Correct Option is B

Step 1: Let \[ y=\frac{x}{x^2-5x+9}. \] We must find all possible values of \(y\).
Cross-multiplying, \[ y(x^2-5x+9)=x. \] \[ yx^2-(5y+1)x+9y=0. \]

Step 2: Apply the condition for real values of \(x\).
For \(x\) to be real, the quadratic equation in \(x\) must have real roots. Hence, \[ \Delta \geq 0. \] Therefore, \[ (5y+1)^2-4(y)(9y)\geq 0. \] \[ 25y^2+10y+1-36y^2\geq 0. \] \[ -11y^2+10y+1\geq 0. \] Multiplying by \(-1\), \[ 11y^2-10y-1\leq 0. \]

Step 3: Solve the quadratic inequality.
Factorizing, \[ 11y^2-10y-1=(11y+1)(y-1). \] Thus, \[ (11y+1)(y-1)\leq 0. \] Hence, \[ -\frac{1}{11}\leq y\leq 1. \]

Step 4: Verify the endpoints.
For \[ y=1, \] \[ x=x^2-5x+9 \] \[ x^2-6x+9=0 \] \[ (x-3)^2=0, \] which gives a real value of \(x\).
For \[ y=-\frac{1}{11}, \] the discriminant becomes zero, so a real value of \(x\) exists.
Hence both endpoints belong to the range.

Step 5: Final conclusion.
Therefore, the range of the function is \[ \boxed{\left[-\frac{1}{11},\,1\right]}. \]