Question

f(X)=
$\frac{(x - 2)(x - 1)}{(x - 3)}$
for all X belongs to 3..find min value of f(X) is equal to ?​

Answer 1

welcome to the concept of inequality and function.

given :

$f(x) = \frac{(x - 2)(x - 1)}{(x - 3)}$

let t = x-3

then ,

x-2 = t +1

x-1 = t +2

then ,

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

$= t + 3 + \frac{2}{t}$

now ,

range of t +(2/t )

by am ,gm inequality

we know that

am >gm

then ,apply it

$t + \frac{2}{t} \geqslant 2 \sqrt{2}$

range is [ 2√2, infinity )...(1)

then range of f(X)

add 3 in the range ..(1)

therefore the range of f(X)

is

$(2 \sqrt{2} + 3 \: to \: infinity)$

⏩answer :

therefore the min of f(x) is

2√2+3

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Answer 2

Answer:

$\frac{(x - 2)(x - 1)}{x - 3} \\ ifx = 3 \\ \frac{(3 - 2)(3 - 1)}{3 - 3 } \\ \frac{1 \times 2}{0} \\ = 0 \\ \\ f(x) = 0$

Answer-0

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