Question

if the domain of function f(x)=x^2-6x+7 is infinity then the range of function is​

Answer 1

Answer:

[-2, infinity]

Answer 2

Given function is,

$\longrightarrow f(x)=x^2-6x+7$

Since coefficient of $x^2$ is positive, the graph of this function represents an upward parabola. Hence it has minima.

To find minima, its derivative should be equated to zero.

$\longrightarrow f'(x)=0$

$\longrightarrow \dfrac{d}{dx}\,\big[x^2-6x+7\big]=0$

$\longrightarrow 2x-6=0$

$\longrightarrow x=3$

So,

$\longrightarrow f(3)=(3)^2-6(3)+7$

$\longrightarrow f(3)=9-18+7$

$\longrightarrow f(3)=-2$

Thus minima of $f(x)$ is at $(3,\ -2).$

$\Longrightarrow f(x)\geq-2$

$\longrightarrow\underline{\underline{f(x)\in[-2,\ \infty)}}$

Therefore, range of $f(x)$ is $[-2,\ \infty).$