Question

Find the ranges of the following functions:

$f\left(x\right)=1-x-x^2$
$f\left(x\right)=\frac{3x+7}{x-8}$
$f\left(x\right)=x^2+2x+8$
$f\left(x\right)=\frac{4x-7}{x-3}$

Answer 1

(i) $f\left(x\right)=1-x-x^2$ has maximum value and it is $\frac{4ac-b^2}{4a}=\frac{-4-1}{-4}=\frac{5}{4}$

Range of $f\left(x\right)=(-\infty ,\frac{5}{4}]$

(ii) $f\left(x\right)=\frac{3x+7}{x-8}$

Let, $y=\frac{3x+7}{x-8}⟹xy-8y=3x+7$

$⟹x(y-3)=8y+7⟹x=\frac{8y+7}{y-3}$

Range of $f\left(x\right)=R-\{3\}$

(iii)$f\left(x\right)=x^2+2x+8$ has minimum value and it is $\frac{4ac-b^2}{4a}=\frac{32-4}{4}=\frac{28}{4}=7$

Range of $f\left(x\right)=[7,\infty )$

(iv) $f\left(x\right)=\frac{3x+7}{x-8}$

Let, $y=\frac{4x-7}{x-3}⟹xy-3y=4x-7$

$⟹x(y-4)=3y-7⟹x=\frac{3y-7}{y-4}$

Range of $f\left(x\right)=R-\{4\}$