Question

for real x, then integer which does not lie in the range of f(x) where f(x) = x^2-2x+2/2x-2

Answer 1

Given:

$\bold{\ f(x) = \frac{x^2-2x+2}{2x-2}}$

To find :

find integer that in not is range.

Solution:

$\ f(x) = \frac{x^2-2x+2}{2x-2} \\\\\ factor \ of \ = x^2-2x+2\\\\ \ compare \ the \ value \\\\ax^2+bx+c=0\\\\\Rightarrow a=1\\ \Rightarrow b=-2\\ \Rightarrow c=2\\\\formula:\\\\x= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\\\x=\frac{2\pm \sqrt{2-8}}{2}\\\\x= \frac{2\pm \sqrt{-6}}{2}\\\\\ f(x) = \frac{\frac{2\pm \sqrt{-6}}{2}}{2x-2} \\\\\ f(x)= \frac{2\pm \sqrt{-6}}{2}} \times \frac{2(x-1)}{1}\\\\\ f(x)= \frac{2\pm \sqrt{-6}}{1}} \times \frac{(x-1)}{1}$

$\Rightarrow \sqrt{6} =2.44\\\\ \Rightarrow f(x) = (2\pm 2.24) (x-1)\\\\ \Rightarrow f(x) = (- 0.24) (x-1)\ \ \ \ and \ \ \ 4.24(x-1)\\\\ \Rightarrow f(x) = -0.24x+0.24 \ \ \ \ and 4.24x-4.24 \\\\if \Rightarrow x = 0 \\\\ \Rightarrow (-0.24x+0.24 ) =0 \ \ \ \ and \ \ \ \(4.24x-4.24)=0$

$\\\\ \Rightarrow -0.24x= -0.24 \ \ \ \ \ \ and \ \ \ \ 4.24x=4.24 \\\\ \Rightarrow 0.24x= 0.24 \ \ \ \ and \ \ \ \ x= \frac{4.24}{4.24}\\\\ \Rightarrow x= \frac{0.24}{0.24} \ \ \ \ and \ \ \ \ x= \frac{4.24}{4.24} \\ \\ \Rightarrow x= 1 \ \ \ \ and \ \ \ x= 1$

The answer is 1, that's why all values in integer.

Answer 2

For real x, the integer which does not lie in the range of f(x) is 0

Step-by-step explanation:

The given function

$f(x)=\frac{x^2-2x+2}{2x-2}$

or, $f(x)=\frac{x^2-2x+2}{2(x-1)}$

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

$\implies f'(x)=\frac{1}{2}\times\frac{2x^2-2x-2x+2-x^2+2x-2}{(x-1)^2}$

$\implies f'(x)=\frac{1}{2}\times\frac{x^2-2x}{(x-1)^2}$

$\implies f'(x)=\frac{x^2-2x}{2(x-1)^2}$

$\implies f'(x)=\frac{x(x-2)}{2(x-1)^2}$

For maxima or minima

$f'(x)=0$

$\implies x(x-2)=0$

$\implies x=0,2$

Taking x = 0 point

If x is slightly less than 0 i.e. x = 0-h = -h

$f'(0-h)=\frac{-h(-h-2)}{2(-h-1)^2}$

or, $f'(0-h)=\frac{h(h+2)}{2(h+1)^2}>0$

If x is slightly greater than 0 i.e. x = 0+h = +h

$f'(0+h)=\frac{h(h-2)}{2(h-1)^2}$

or, $f'(0-h)=\frac{h(h-2)}{2(h+1)^2}<0$

i.e. at x=0 there is local maxima

Similarly we can show that at x = 2 there is local minima

At x = 0

$f(x)=-1$

At x = 2

$f(x)=1$

At x = 1, the value tends towards infinity

As x → 1⁻

f(x) → -∞

As x → 1⁺

f(x) → +∞

Also,

$f(x)=\frac{(x-1)^2+1}{2(x-1)}$

$\implies f(x)=\frac{x-1}{2}+\frac{1}{2(x-1)}$

As x → -∞

f(x) → -∞

As x → +∞

f(x) → +∞

Thus, the graph of f(x)  will look like the figure attached.

From graph it is clear that the only integer that isn't in the range of f(x) is zero

Hope this answer is helpful.

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