Question

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$f(x) = \dfrac{1}{ {x}^{2} - x + 1 }$
Find range of the function.​

Answer 1

$\huge{ \rm\left(0 , \dfrac{4}{3} \right ]}$

$\Large{\underline{\underline{\text{Explanation:}}}}$

Method 1:

For any polynomial of the form $\rm ax^2 + bx + c$ where $\rm a > 0$, it's range is given by $\rm \left[\dfrac{-D}{4a} , \infty\right)$ where $\rm D = b^2 - 4ac$ or simply the discriminant.

Consider the denominator of given function,

$\longrightarrow \rm {x}^{2} - x + 1$

Here,

• a = 1
• b = -1
• c = 1

Finding determinant:

$\cdot \cdot \cdot \longrightarrow \rm D = ( - 1)^{2} - 4(1)(1)$

$\cdot \cdot \cdot \longrightarrow \rm D = 1 - 4$

$\cdot \cdot \cdot \longrightarrow \rm D = - 3$

So, the range of denominator is given by,

$\implies \rm\left[\dfrac{-D}{4a} , \infty\right)$

$\implies \rm\left[\dfrac{-( - 3)}{4(1)} , \infty\right)$

$\implies \rm\left[\dfrac{3}{4} , \infty\right)$

Range of $\rm x^2 - x + 1$ is $\rm\left[\dfrac{3}{4} , \infty\right)$ therefore the range of given function is $\rm \left(0 , \dfrac{4}{3} \right ]$.

Method 2:

Consider the denominator of given function,

$\longrightarrow \rm {x}^{2} - x + 1$

Using completing the square,

$\longrightarrow \rm\left( x - \dfrac{1}{2} \right)^2 + \dfrac{3}{4}$

So clearly it's range is $\left[\dfrac{3}{4} , \infty\right)$. and hence the range of original function is $\rm \left(0 , \dfrac{4}{3} \right ]$

$\Large{\underline{\underline{\text{Note that:}}}}$

Since the denominator is ranging from $\left[ \dfrac{3}{4} , \infty\right)$ and the numerator is $1$, this gives us following results:

• If denominator is $\rm \infty , \ f(x) \to 0$
• If denominator is $\rm \dfrac{3}{4}, \ f(x) = \dfrac{4}{3}$.

$\Large{\underline{\underline{\text{Answer:}}}}$

$\rm 0 < f(x) \leqslant \dfrac{4}{3}$

Answer 2

Answer:

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