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Differential Calculus

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18 Jan 2012 12:10:01 IST

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Find the domain of underout X^12-X^9+X^4-X+1

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Find the domain of underout X^12-X^9+X^4-X+1

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jagdish singh

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18 Jan 2012 13:11:57 IST

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$\hspace{-16}\mathbf{(1)\;\;:: f(x)=\sqrt{x^{12}-x^9+x^4-x+1}}$\\\\ Now function $\mathbf{f(x)}$ is defined only when $\mathbf{x^{12}-x^9+x^4-x+1>=0}$\\\\ Now Let $\mathbf{g(x)=x^{12}-x^9+x^4-x+1}$\\\\ \textbf{case{(a)::\;}}If $\mathbf{x\leq 0\;,}$ Then $\mathbf{g(x)>0}$\\\\ \textbf{case{(b)::\;}}If $\mathbf{0<x<1 \;,}$ Then $\mathbf{g(x)=x^{12}+x^4-x^9+1-x=x^{12}+x^4(1-x^5)+(1-x)>0}$\\\\ \textbf{case{(c)::\;}}If $\mathbf{x\geq 1 \;,}$ Then $\mathbf{g(x)=x^{12}-x^9+x^4-x+1=x^{9}(x^3-1)+x(x^3-1)+1>0}$\\\\ So $\mathbf{g(x)>0\forall x\in\mathbb{R}}$\\\\ So Domain is $\mathbf{x\in\mathbb{R}}$$

ROHAN KUMAR

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19 Jan 2012 00:18:23 IST

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how u know that if x<_0. then g(x)>0...just explain me

Abhishek Bandiya

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19 Jan 2012 04:13:08 IST

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JAGDISH IS RIGHT

if x<_0 than g(X)>0

suppose y is a positive quantity or equal to zero and we have to put x<_0

then we will put x=-y =-ve quantity

g(-y)=

if even y=0 g(-y) =1>0

g(-y)>0 i.e.

g(x)>0 where x<_0

Arjun Virmani

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24 Jan 2012 18:12:11 IST

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the only constrain to (sq root) is negative values......

The quantity under the root can't be -ve.

for x <= 0

x^9 ( x^3 - 1) + x ( x^3 -1) + 1

(x^9 + x) * ( x^3 -1) + 1

(x^9 + x) is -ve for all x<0

and ( x^3 -1) is also -ve for all x<0....

-ve * -ve = +ve .......... (+ 1) = +ve.

thrfr the f is >0 for all x<=0................!!!!!

ly for all x>1

(x^9 + x) and ( x^3 -1) are +ve thrfr the function is +ve!!!

and ly for remaining x also!!!

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