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Best Answer:
Given q^2-4pr>0 & p>0.

f(x)=log[px^3+(p+q)x^2+(q+r)x+r]

=>

f(x)=log[(x+1)(px^2+qx+r)]

=>

f(x)=log{p(x+1)[(x+q/(2p))^2-

[sqr(q^2-4pr)/(2p)]^2]}

(completing the square)

=>

f(x)=log{p(x+1)[(x+q/(2p))^2-

[sqr(D)/(2p)]^2]}

(where D=q^2-4pr)

=>

Dom(f(x))

=

(x>-1) & [{(x+q/(2p))^2-[sqr(D)/(2p)]^2}>0]

=

(x>-1) & [(x<-(sqr(D)+q)/(2p)) or (x>(sqr(D)-q)/(2p)]

=

(x>-1) & (x<-(sqr(D)+q)/(2p)

or

(x>-1) & (x>(sqr(D)-q)/(2p)

f(x)=log[px^3+(p+q)x^2+(q+r)x+r]

=>

f(x)=log[(x+1)(px^2+qx+r)]

=>

f(x)=log{p(x+1)[(x+q/(2p))^2-

[sqr(q^2-4pr)/(2p)]^2]}

(completing the square)

=>

f(x)=log{p(x+1)[(x+q/(2p))^2-

[sqr(D)/(2p)]^2]}

(where D=q^2-4pr)

=>

Dom(f(x))

=

(x>-1) & [{(x+q/(2p))^2-[sqr(D)/(2p)]^2}>0]

=

(x>-1) & [(x<-(sqr(D)+q)/(2p)) or (x>(sqr(D)-q)/(2p)]

=

(x>-1) & (x<-(sqr(D)+q)/(2p)

or

(x>-1) & (x>(sqr(D)-q)/(2p)

Pinkgreen
· 3 weeks ago

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