IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

reddevil_2009

Find domain:

$\sqrt{(x}$ ¹² - x⁹+ x⁴ -x +1)

Find range:

1) (x² +2x + 3)/3

2) $\sqrt{x - 1}$ + $\sqrt{5 - x}$

bumba

Range::::-

1.[2/3,infinity)

solution::

(x² +2x + 3)/3

=x² +2x + 1+2/3

=(x+1)² +2/3

so,only when the value of x=-1, we get the minimum valu of the expression to be 2/3 and the maximum valu will be tending to infinity.

thus,range is [2/3,infinity)

2.range::[2,2 $\sqrt{2}$ ]

shinee

can u tell me how
(x^2 +2x + 3)/3 =x^2 +2x + 1+2/3

ashish_banga

hey Bumba it is ( x+ 1 )^2 / 3
answer is perfect

Aatish

@ shreya.....just apply proper brackets.....and write 3 = 1 + 2.......

u will get it.......ok........and bumba is correct////

shinee

ok fine after applying the brackets, the expression becomes
(3x^2+6x+5)/3 and how is this equal to (x^2 +2x + 3)/3??

Aatish

$\frac{{x}^{2} + 2x + 3 }{3}$

can be rewritten as..........

$\frac{({x}^{2}+2x+1)+2}{3}$

now do u have any problem!!!!!!!!!

joyfrancis

x^2 + 2x + 3 is an upward parabola with imaginary roots, so it is always positive , it's least value is -D/4a
= -(-8)/4 = 2..so range of the function (x^2 + 2x + 3)/3 is [2/3,inf)

joyfrancis

for the last one , just diff the eqn once..

critical pts come out to be x=1,3,5 at x=1,5 f(x)=2 and at x=3 f(x)=2root2 so range = [2,2root2]

reddevil_2009

solve the first one tooooooooooooooooooooooo

spideyunlimited

$\sqrt{x^{12} - x^9 + x^4 - x +1 }$

= $\sqrt{x^{12} + x^4 - (x^9 + x) + 1}$

= $\sqrt{ (x^4 - x)(x^8 + 1) + 1 }$

x⁸ + 1 will always be positive.

x⁴ - x will be least at x = (1/4)^1/3 ...............found by diff. and equating to 0.

But at this value, the expression under the root is greater than 0.

Hence domain is ALL REAL NUMBERS.

Greatdreams

${x}^{12}+{x}^{9}+{x}^{4}-{x} + 1 \; is\; defined\; for > 0$

$Hence \; {x}^{4}({x}^{8}\; + 1) - x ({x}^8 + 1) + 1 > 0$

$So , ({x}^{8} +1)x \; ({x}^{3} - 1) + 1 > 0$

$If \;x\ge1 \;or \le -1 , given\; expression\; assumes \;positive\; values$

$If \;-1<x\le0 \; then \;again \;expression \;is > 0$

$Considering \; 0<x<1\; inequality \;still \;remains \;true$

$So\;domain \;(-\infty , \infty)$

$sorry \;a \;little\; bit \;edition :$ $f(x) \; is\; defined \;for \;the \;expression\; being \;\ge and\; not \;> 0$

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