Question #248725

Precalculus

Good day sir could you please assist for solutions to this problems for Precalculus Pre-engineering, which is due today.


*Problem 1*


For the expression

f(x)=log((1-|x|) ÷ (1+|x|))


Determine:


• it's largest domain;

• it's intersections with the x-axis (if any);

• it's intersection with the y-axis (if any);

• it's sign;

• when appropriate, end behaviours and behaviours at accumulation points of the domain which are not in the domain, possible symptoms.


*Problem 2*

Compute, showing the procedure,


limit from x to positive infinity ( square root symbol with x^2 - 3x - square root symbol with x^2 - 5x + 1)


*Problem 3*


Compute, showing the procedure.,


limit from x to negative infinity of the expression (square root with x^6 - x^2)÷(1-2x)


1
Expert's answer
2021-10-12T01:54:18-0400

Problem 1


f(x)=log(1x1+x)f(x)=\log(\dfrac{1-|x|}{1+|x|})

1x1+x>0=>1x>0=>x<1\dfrac{1-|x|}{1+|x|}>0=>1-|x|>0=>|x|<1

Domain:(1,1)Domain: (-1, 1)

xx- intercept: y=0=>0=log(1x1+x)=>1x1+x=1y=0=>0=\log(\dfrac{1-|x|}{1+|x|})=>\dfrac{1-|x|}{1+|x|}=1



=>1x=1+x=>x=0=>1-|x|=1+|x|=>x=0

Point (0,0).(0, 0).

yy- intercept: x=0=>y(0)=log(101+0)=0x=0=>y(0)=\log(\dfrac{1-|0|}{1+|0|})=0

Point (0,0).(0, 0).

The graph passes through the origin.


1x1+x=1,x=0\dfrac{1-|x|}{1+|x|}=1, x=0

0<1x1+x<1,x(1,0)(0,1)0<\dfrac{1-|x|}{1+|x|}<1, x\in(-1, 0)\cup(0, 1)

Then f(x)<0,x(1,0)(0,1)f(x)<0, x\in(-1, 0)\cup(0, 1) and f(0)=0.f(0)=0.

Range:(,)Range: (-\infin, \infin)


limx1+f(x)=limx1+log(1x1+x)=\lim\limits_{x\to-1^+}f(x)=\lim\limits_{x\to-1^+}\log(\dfrac{1-|x|}{1+|x|})=-\infin

limx1f(x)=limx1log(1x1+x)=\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}\log(\dfrac{1-|x|}{1+|x|})=-\infin



Problem 2


limx(x23xx25x+1)\lim\limits_{x\to\infin}(\sqrt{x^2-3x}-\sqrt{x^2-5x+1})

=limx(x23x(x25x+1)x23x+x25x+1)=\lim\limits_{x\to\infin}(\dfrac{x^2-3x-(x^2-5x+1)}{\sqrt{x^2-3x}+\sqrt{x^2-5x+1}})

=limx(2xx1xx2x23xx2+x2x25xx2+1x2)=\lim\limits_{x\to\infin}(\dfrac{\dfrac{2x}{x}-\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{3x}{x^2}}+\sqrt{\dfrac{x^2}{x^2}-\dfrac{5x}{x^2}+\dfrac{1}{x^2}}})

=limx(21x13x+15x+1x2)=\lim\limits_{x\to\infin}(\dfrac{2-\dfrac{1}{x}}{\sqrt{1-\dfrac{3}{x}}+\sqrt{1-\dfrac{5}{x}+\dfrac{1}{x^2}}})

=2010+10+0=1=\dfrac{2-0}{\sqrt{1-0}+\sqrt{1-0+0}}=1

Problem 3


limxx6x212x\lim\limits_{x\to-\infin}\dfrac{\sqrt{x^6-x^2}}{1-2x}

=limxxx4112x=\lim\limits_{x\to-\infin}\dfrac{-x\sqrt{x^4-1}}{1-2x}

=limxx411x2xx=\lim\limits_{x\to-\infin}\dfrac{-\sqrt{x^4-1}}{\dfrac{1}{x}-\dfrac{2x}{x}}

=limxx4121x==\lim\limits_{x\to-\infin}\dfrac{\sqrt{x^4-1}}{2-\dfrac{1}{x}}=\infin


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS