Answer to Question #248725 in Mechanical Engineering for Badge

Question

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)

Answer 1

Problem 1

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

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

$Domain: (-1, 1)$

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

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

Point $(0, 0).$

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

Point $(0, 0).$

The graph passes through the origin.

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

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

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

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

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

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

Problem 2

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

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

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

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

$$=\dfrac{2-0}{\sqrt{1-0}+\sqrt{1-0+0}}=1$$

Problem 3

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

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

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

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