Question

find the
range of the above question​

Answer 1

Given :

• f ( x ) = $\sqrt {x^2 + x + 1 }$

To find :

• Range of the function of f ( x ) = $\sqrt {x^2 + x + 1 }$

According to the question :

➳ f ( x ) = $\sqrt {x^2 + x + 1 }$

➳ f ( x ) = $\sqrt{x + 1 / 2^2 + 3 / 4}$

➳ Clearly range of f ( x ) = $\frac {\sqrt{3}} 2 , ∞$

➳ Minimum occurs at x = $-1 / 2$

More information :

↦Solve the equation for ' x ' Eg: x = g ( y )

↦Find the domain of g ( y ) , and this will be the range of f ( x )

↦If you can't seem to find the ' x ' , then try graphing the function to find ' x '.

So, It's Done !!

Answer 2

Given:

$\sf \bull f(x) = \sqrt{ {x}^{2} + x + 1 }$

Find:

$\sf \bull range \: of \:function \: of \: f(x) = \sqrt{ {x}^{2} + x + 1 }$

Solution:

$\sf Here, f(x) = \sqrt{ {x}^{2} + x + 1 }$

$\sf \looparrowright f(x) = \sqrt{ {x}^{2} + x + 1 }$

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

$\sf \implies f(x) = \sqrt{ {( x + \frac{1}{2}) }^{2} + \frac{3}{4} \geqslant \sqrt{ \frac{3}{4} } }$

$\sf \implies f(x) \geqslant \frac{ \sqrt{3} }{2}$

$\sf Hence, range \: of \: f(x) = ( \frac{ \sqrt{3} }{2} , \infin )$

$\sf Minimum \: occurs \: at \: x = \frac{ - 1}{2}$