Question

the range of the function f(x)=x3-8/x-2

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x)=\dfrac{x^3-8}{x-2}}$

Now,

$\sf{Now,\,\,let\,\,f(x)=y=\dfrac{x^3-8}{x-2}}$

$\sf{\implies\,y(x-2)=(x-2)(x^2-2x+4)}$

$\sf{\implies\,y(x-2)-(x-2)(x^2-2x+4)=0}$

$\sf{\implies\,(x-2)\{y-(x^2-2x+4)\}=0}$

$\sf{\implies\,(x-2)\{y-x^2+2x-4\}=0}$

$\sf{\implies\,x^2-2x+4-y=0}$

$\bf{Now,\,\,its \,\,discriminant\,\,will\,\,positive}$

$\tt{(-2)^2-4(4-y)(1)>0}$

$\tt{\implies4-4(4-y)>0}$

$\tt{\implies4\{1-(4-y)\}>0}$

$\tt{\implies1-(4-y)>0}$

$\tt{\implies1-4+y>0}$

$\tt{\implies\,y-3>0}$

$\tt{\implies\,y>3}$

Hence, range of the given function $\sf{\in(3,\infty)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{ {x}^{3} - 8}{x - 2}$

To find the range of the function, Let assume that

$\rm :\longmapsto\:y = \dfrac{ {x}^{3} - 8}{x - 2}$

$\rm :\longmapsto\:y = \dfrac{ {x}^{3} - {2}^{3} }{x - 2}$

We know,

$\red{\rm :\longmapsto\: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }$

So, using this, we get

$\rm :\longmapsto\:y = \dfrac{ (x - 2)( {x}^{2} + 2x + {2}^{2} )}{x - 2}$

$\rm \implies\:y = {x}^{2} + 2x + 4$

$\rm \implies\:{x}^{2} + 2x + 4 - y = 0$

For roots to be real, Discriminant must be greater than or equals to 0.

$\rm \implies\: {2}^{2} - 4 \times 1 \times (4 - y) \geqslant 0$

$\rm \implies\: 4 - 4 (4 - y) \geqslant 0$

$\rm \implies\: 4 [1 - (4 - y)] \geqslant 0$

$\rm \implies\: 1 - 4 + y\geqslant 0$

$\rm \implies\: - 3 + y\geqslant 0$

$\rm \implies\: y\geqslant 3$

$\bf\implies \:y \: \in \: [3, \: \infty )$

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Basic Concept Used :-

To find the range of function f(x)

Write down y=f(x).

Solve the equation for x to represent in terms of y

Represent in the form x = g(y).

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