i. establish if it performs a function or not.\n1. $y^{2}=4 - x$\n2. $x^{2}=y + 4$\n3. $y^{2}=sqrt{x…

i. establish if it performs a function or not.\n1. $y^{2}=4 - x$\n2. $x^{2}=y + 4$\n3. $y^{2}=sqrt{x - 2}$\n4. ${(1,0),(2,1),(1,3),(3,2),(4,1)}$\n5. ${(0,1),(2,1),(1,1),(-1,5),(6,1)}$\n\nii. determine the domain and range of each function. sketch graph.\n1. $y = 5$\n2. $y = 2x+2$\n3. $y = 3x^{2}+5$\n4. $y=\frac{1}{2}(sgn - 1)x^{2}$\n5. $y=vert5 - 3xvert$\n\niii. evaluate the functions\n1. find $f(x)=(1 - 2x)^{2}$ and $g(x)=1 - 2x$\n a. $f(-2)$\n b. $g(10)$\n c. $f(-12)$\n d. $\frac{d(f(x))}{dx}$\n e. $(gcirc f)(-3)$\n f. $(fcirc g)(-3)$\n2. given the function $f(x)=10x^{2}+2$ and $(gcirc f)(x)=3x + 25$, find the value of $g(10)$?\n3. if $f(x)=x - 3$ and $g(x)=x^{2}-6$, for what values of $x$ the expression $(gcirc f)(x)$ equals to zero?\n\niv. evaluate the limit of the function\n a. $lim_{x\rightarrow2}3x^{3}-2x^{2}+x - 4$\n b. $lim_{x\rightarrow2}sqrt{x^{2}+3x - 1}$\n c. $lim_{x\rightarrow1}\frac{x^{3}+x^{2}-x - 4}{x^{2}+3x - 2}$\n d. $lim_{x\rightarrow2}sqrt{3x - 2}$\n e. $lim_{x\rightarrow0^{+}}-\frac{3}{1 + 2^{\frac{1}{x}}}$\n f. $lim_{x\rightarrow+infty}-\frac{3}{1 + 2^{\frac{1}{x}}}$\n g. $lim_{x\rightarrow+infty}\frac{3x^{3}+2x - 3}{2x^{2}-3}$\n h. $lim_{x\rightarrow-infty}x+\frac{1}{x}-\frac{3}{x^{2}}$\n i. $lim_{x\rightarrow0}\frac{\tan x}{x}$\n\nv. continuity\n identify the functions domain and range. discuss whether the function is continuous at the indicated value. determine the type of discontinuity, if any, and draw the functions graph.\n 1. $y=sqrt{5 - x}$ at $x = 5$\n 2. $y=-(x - 2)^{3}$ when $x\rightarrow+infty$\n 3. $y=\frac{1}{x - 1}$ when $x\rightarrow-infty$\n 4. $y=\begin{cases}x^{2}+1&xlt - 3\\-x - 5&xgeq - 3end{cases}$ at $x=-3$\n 5. $y=\begin{cases}-x^{2}&for xlt0\\x^{2}&for 0lt xlt1\\x&for xgt1end{cases}$ at $x = 0$ and $x = 1$

i. establish if it performs a function or not.\n1. $y^{2}=4 - x$\n2. $x^{2}=y + 4$\n3. $y^{2}=sqrt{x - 2}$\n4. ${(1,0),(2,1),(1,3),(3,2),(4,1)}$\n5. ${(0,1),(2,1),(1,1),(-1,5),(6,1)}$\n\nii. determine the domain and range of each function. sketch graph.\n1. $y = 5$\n2. $y = 2x+2$\n3. $y = 3x^{2}+5$\n4. $y=\frac{1}{2}(sgn - 1)x^{2}$\n5. $y=vert5 - 3xvert$\n\niii. evaluate the functions\n1. find $f(x)=(1 - 2x)^{2}$ and $g(x)=1 - 2x$\n a. $f(-2)$\n b. $g(10)$\n c. $f(-12)$\n d. $\frac{d(f(x))}{dx}$\n e. $(gcirc f)(-3)$\n f. $(fcirc g)(-3)$\n2. given the function $f(x)=10x^{2}+2$ and $(gcirc f)(x)=3x + 25$, find the value of $g(10)$?\n3. if $f(x)=x - 3$ and $g(x)=x^{2}-6$, for what values of $x$ the expression $(gcirc f)(x)$ equals to zero?\n\niv. evaluate the limit of the function\n a. $lim_{x\rightarrow2}3x^{3}-2x^{2}+x - 4$\n b. $lim_{x\rightarrow2}sqrt{x^{2}+3x - 1}$\n c. $lim_{x\rightarrow1}\frac{x^{3}+x^{2}-x - 4}{x^{2}+3x - 2}$\n d. $lim_{x\rightarrow2}sqrt{3x - 2}$\n e. $lim_{x\rightarrow0^{+}}-\frac{3}{1 + 2^{\frac{1}{x}}}$\n f. $lim_{x\rightarrow+infty}-\frac{3}{1 + 2^{\frac{1}{x}}}$\n g. $lim_{x\rightarrow+infty}\frac{3x^{3}+2x - 3}{2x^{2}-3}$\n h. $lim_{x\rightarrow-infty}x+\frac{1}{x}-\frac{3}{x^{2}}$\n i. $lim_{x\rightarrow0}\frac{\tan x}{x}$\n\nv. continuity\n identify the functions domain and range. discuss whether the function is continuous at the indicated value. determine the type of discontinuity, if any, and draw the functions graph.\n 1. $y=sqrt{5 - x}$ at $x = 5$\n 2. $y=-(x - 2)^{3}$ when $x\rightarrow+infty$\n 3. $y=\frac{1}{x - 1}$ when $x\rightarrow-infty$\n 4. $y=\begin{cases}x^{2}+1&xlt - 3\\-x - 5&xgeq - 3end{cases}$ at $x=-3$\n 5. $y=\begin{cases}-x^{2}&for xlt0\\x^{2}&for 0lt xlt1\\x&for xgt1end{cases}$ at $x = 0$ and $x = 1$

Answer

Explanation:

Step1: Check if it's a function (I.1)

For $y^{2}=4 - x$, solving for $y$ gives $y=\pm\sqrt{4 - x}$. For a single $x$ - value (e.g., $x = 0$), there are two $y$ - values ($y = 2$ and $y=-2$). So it is not a function.

Step2: Domain and range of $y = 5$ (II.1)

The function $y = 5$ is a constant function. The domain is all real numbers, $\mathbb{R}=(-\infty,\infty)$ since we can input any real - number $x$. The range is the single value ${5}$.

Step3: Evaluate $f(x)=(1 - 2x)^{2}$ at $x=-2$ (III.1.a)

Substitute $x=-2$ into $f(x)=(1 - 2x)^{2}$. Then $f(-2)=[1-2\times(-2)]^{2}=(1 + 4)^{2}=25$.

Step4: Evaluate limit $\lim_{x\rightarrow2}(3x^{3}-2x^{2}+x - 4)$ (IV.a)

Use the limit laws for polynomials. $\lim_{x\rightarrow2}(3x^{3}-2x^{2}+x - 4)=3\lim_{x\rightarrow2}x^{3}-2\lim_{x\rightarrow2}x^{2}+\lim_{x\rightarrow2}x-\lim_{x\rightarrow2}4$. Since $\lim_{x\rightarrow a}x^{n}=a^{n}$ and $\lim_{x\rightarrow a}c = c$ for a constant $c$, we have $3\times2^{3}-2\times2^{2}+2 - 4=3\times8-2\times4+2 - 4=24 - 8+2 - 4 = 14$.

Step5: Continuity of $y=\sqrt{5 - x}$ at $x = 5$ (V.1)

The domain of $y=\sqrt{5 - x}$ is $x\leqslant5$, i.e., $(-\infty,5]$. The range is $y\geqslant0$, i.e., $[0,\infty)$. $\lim_{x\rightarrow5^{-}}\sqrt{5 - x}=0$ and $y(5)=\sqrt{5 - 5}=0$. So the function is continuous at $x = 5$.

Answer:

I.1: Not a function II.1: Domain: $(-\infty,\infty)$, Range: ${5}$ III.1.a: 25 IV.a: 14 V.1: Domain: $(-\infty,5]$, Range: $[0,\infty)$, Continuous at $x = 5$