directions: for each of the following, write the left and right limit statements for f(x) as x approaches…

directions: for each of the following, write the left and right limit statements for f(x) as x approaches 1.\n17. f(x) = \\frac{(x - 1)(x + 5)}{(x - 1)(x + 2)}\nleft:\nright:\n18. f(x) = \\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}\nleft:\nright:\n19. f(x) = \\frac{-2(x + 3)(x + 1)}{(x - 1)^2}\nleft:\nright:\ndirections: for each of the following rational functions, determine and label any values of x where the graph has a hole or vertical asymptote.\n20. y = \\frac{(x + 3)(x - 2)}{(x + 3)^2(x - 2)}\n21. k(x) = \\frac{(x + 7)(x + 2)^3}{(x + 1)(x + 2)^2}\n22. r(x) = \\frac{x^3 - x^2}{x^2 + 2x + 1}\ndirections: solve the following inequalities. write your answers using interval notation.\n23. \\frac{x - 3}{x + 2} \\leq 0\n24. \\frac{(x - 1)^2(x + 2)}{(x + 1)}>0\n25. \\frac{x^2 - x - 12}{x^2} \\geq 0\n26. \\frac{-2x(x - 3)^2}{(x + 2)(x - 4)^2}<0

directions: for each of the following, write the left and right limit statements for f(x) as x approaches 1.\n17. f(x) = \\frac{(x - 1)(x + 5)}{(x - 1)(x + 2)}\nleft:\nright:\n18. f(x) = \\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}\nleft:\nright:\n19. f(x) = \\frac{-2(x + 3)(x + 1)}{(x - 1)^2}\nleft:\nright:\ndirections: for each of the following rational functions, determine and label any values of x where the graph has a hole or vertical asymptote.\n20. y = \\frac{(x + 3)(x - 2)}{(x + 3)^2(x - 2)}\n21. k(x) = \\frac{(x + 7)(x + 2)^3}{(x + 1)(x + 2)^2}\n22. r(x) = \\frac{x^3 - x^2}{x^2 + 2x + 1}\ndirections: solve the following inequalities. write your answers using interval notation.\n23. \\frac{x - 3}{x + 2} \\leq 0\n24. \\frac{(x - 1)^2(x + 2)}{(x + 1)}>0\n25. \\frac{x^2 - x - 12}{x^2} \\geq 0\n26. \\frac{-2x(x - 3)^2}{(x + 2)(x - 4)^2}<0

Answer

Problem 17:

Explanation:

Step1: Simplify the function

$f(x)=\frac{(x - 1)(x + 5)}{(x - 1)(x + 2)}=\frac{x + 5}{x + 2},x\neq1$

Step2: Find the left - hand limit as $x\to1$

$\lim_{x\to1^{-}}\frac{x + 5}{x + 2}=\frac{1+5}{1 + 2}=2$

Step3: Find the right - hand limit as $x\to1$

$\lim_{x\to1^{+}}\frac{x + 5}{x + 2}=\frac{1+5}{1 + 2}=2$

Answer:

Left: $\lim_{x\to1^{-}}f(x)=2$ Right: $\lim_{x\to1^{+}}f(x)=2$

Problem 18:

Explanation:

Step1: Simplify the function

$f(x)=\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}$, no simplification possible.

Step2: Find the left - hand limit as $x\to1$

$\lim_{x\to1^{-}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}$, as $x\to1^{-}$, $(x - 2)(x - 4)\to(-1)\times(-3) = 3$ and $(x - 1)(x + 2)\to0^{-}$, so $\lim_{x\to1^{-}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}=-\infty$

Step3: Find the right - hand limit as $x\to1$

$\lim_{x\to1^{+}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}$, as $x\to1^{+}$, $(x - 2)(x - 4)\to(-1)\times(-3)=3$ and $(x - 1)(x + 2)\to0^{+}$, so $\lim_{x\to1^{+}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}=\infty$

Answer:

Left: $\lim_{x\to1^{-}}f(x)=-\infty$ Right: $\lim_{x\to1^{+}}f(x)=\infty$

Problem 19:

Explanation:

Step1: Analyze the function

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

Step2: Find the left - hand limit as $x\to1$

As $x\to1^{-}$, $-2(x + 3)(x + 1)\to-2\times4\times2=-16$ and $(x - 1)^{2}\to0^{+}$, so $\lim_{x\to1^{-}}f(x)=-\infty$

Step3: Find the right - hand limit as $x\to1$

As $x\to1^{+}$, $-2(x + 3)(x + 1)\to-2\times4\times2=-16$ and $(x - 1)^{2}\to0^{+}$, so $\lim_{x\to1^{+}}f(x)=-\infty$

Answer:

Left: $\lim_{x\to1^{-}}f(x)=-\infty$ Right: $\lim_{x\to1^{+}}f(x)=-\infty$

Problem 20:

Explanation:

Step1: Simplify the function

$y=\frac{(x + 3)(x - 2)}{(x + 3)^{2}(x - 2)}=\frac{1}{x + 3},x\neq - 3,x\neq2$

Step2: Find holes and asymptotes

The function has a hole at $x = 2$ (since the factor $(x - 2)$ cancels) and a vertical asymptote at $x=-3$

Answer:

Hole at $x = 2$, Vertical asymptote at $x=-3$

Problem 21:

Explanation:

Step1: Simplify the function

$k(x)=\frac{(x + 7)(x + 2)^{3}}{(x + 1)(x + 2)^{2}}=\frac{(x + 7)(x + 2)}{x + 1},x\neq - 2$

Step2: Find holes and asymptotes

The function has a hole at $x=-2$ and a vertical asymptote at $x=-1$

Answer:

Hole at $x=-2$, Vertical asymptote at $x=-1$

Problem 22:

Explanation:

Step1: Factor the functions

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

Step2: Find holes and asymptotes

There are no common factors in the numerator and denominator that can be canceled, so there are no holes. The vertical asymptote is at $x=-1$

Answer:

No hole, Vertical asymptote at $x=-1$

Problem 23:

Explanation:

Step1: Find the critical points

The critical points of $\frac{x - 3}{x+2}\leq0$ are $x = 3$ and $x=-2$ (where the numerator and denominator are zero respectively).

Step2: Test intervals

Test the intervals $(-\infty,-2)$, $(-2,3]$ and $(3,\infty)$. For $x\in(-\infty,-2)$, let $x=-3$, then $\frac{-3 - 3}{-3+2}=6>0$. For $x\in(-2,3]$, let $x = 0$, then $\frac{0 - 3}{0+2}=-\frac{3}{2}\leq0$.

Answer:

$(-2,3]$

Problem 24:

Explanation:

Step1: Find the critical points

The critical points of $\frac{(x - 1)^{2}(x + 2)}{x + 1}>0$ are $x=-2,x=-1$ and $x = 1$.

Step2: Test intervals

Test the intervals $(-\infty,-2)$, $(-2,-1)$, $(-1,1)$ and $(1,\infty)$. The function is non - negative for $x\in(-\infty,-2)\cup(-1,\infty)$

Answer:

$(-\infty,-2)\cup(-1,\infty)$

Problem 25:

Explanation:

Step1: Factor the numerator

$\frac{x^{2}-x - 12}{x^{2}}=\frac{(x - 4)(x+3)}{x^{2}}\geq0$. The critical points are $x=-3,x = 0$ and $x = 4$.

Step2: Test intervals

Test the intervals $(-\infty,-3]$, $(-3,0)$, $(0,4]$ and $(4,\infty)$. The solution is $(-\infty,-3]\cup[4,\infty)$

Answer:

$(-\infty,-3]\cup[4,\infty)$

Problem 26:

Explanation:

Step1: Find the critical points

The critical points of $\frac{-2x(x - 3)^{2}}{(x + 2)(x - 4)^{2}}<0$ are $x=-2,x = 0,x = 3$ and $x = 4$.

Step2: Test intervals

Test the intervals $(-\infty,-2)$, $(-2,0)$, $(0,3)$, $(3,4)$ and $(4,\infty)$. The solution is $(-2,0)\cup(3,4)\cup(4,\infty)$

Answer:

$(-2,0)\cup(3,4)\cup(4,\infty)$