For what values of $x$ will the expression represent a real number?
Write answers using inequality notation.
$\sqrt{2 x^{2} + x - 6}$

Open in App

Solution

$D o m a i n o f \sqrt{2 x^{2} + x - 6} : x \leq - 2 o r x \geq \frac{3}{2}$

$R a n g e o f \sqrt{2 x^{2} + x - 6} : [\begin{matrix} S o l u t i o n : & f (x) \geq 0 I n t e r v a l N o t a t i o n : & [0, \infty) \end{matrix}]$

$f^{'} (x) = \frac{4 x + 1}{2 \sqrt{2 x^{2} + x - 6}}$

$T h e c r i t i c a l p o i n t s : x = - 2, x = \frac{3}{2}$

$E x t r e m e P o i n t s o f \sqrt{2 x^{2} + x - 6} : M i n i m u m (- 2, 0), M i n i m u m (\frac{3}{2}, 0)$

$T h e r a n g e f o r t h e i n t e r v a l - \infty < x \leq - 2 : 0 \leq f (x) < \infty$

$T h e r a n g e f o r t h e i n t e r v a l \frac{3}{2} \leq x < \infty : 0 \leq f (x) < \infty$

Suggest Corrections

Similar questions

x

\sqrt{3 x^{2} - 7 x - 6}

x

\sqrt{\frac{x - 1}{x + 3}}

x^{2} + x < 12

\frac{x + 4}{1 - x} \leq 0

x^{2} > 4

Join BYJU'S Learning Program