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f(x,y)=4xy−x4−y4 f(x, y)=4 x y-x^{4}-y^{4} \newlineSolution. Since\newlinefx(x,y)=4y−4x3fy(x,y)=4x−4y3 \begin{array}{l} f_{x}(x, y)=4 y-4 x^{3} \\ f_{y}(x, y)=4 x-4 y^{3} \end{array} \newlinethe critical points of f f have coordinates satisfying the equations\newline4y−4x3=0y=x34x−4y3=0 or =y3 \begin{array}{lll} 4 y-4 x^{3}=0 & & y=x^{3} \\ 4 x-4 y^{3}=0 \end{array} \quad \text { or } \quad \begin{array}{ll} & =y^{3} \end{array} \newlineSubstituting the top equation in the bottom yields x=(x3)3 x=\left(x^{3}\right)^{3} or x9−x=0 x^{9}-x=0 o x(x8−1)=0 x\left(x^{8}-1\right)=0 , which has solutions x=0,x=1,x=−1 x=0, x=1, x=-1 . Substituting these values in th top equation of (22) we obtain the corresponding y y values y=0,y=1,y=−1 y=0, y=1, y=-1 . Thus, th critical points of f f are (0,0),(1,1) (0,0),(1,1) , and (−1,−1) (-1,-1) .\newlineFrom x=(x3)3 x=\left(x^{3}\right)^{3} 00\newlinefxx(x,y)=−12x2,fyy(x,y)=−12y2,fxy(x,y)=4 f_{x x}(x, y)=-12 x^{2}, \quad f_{y y}(x, y)=-12 y^{2}, \quad f_{x y}(x, y)=4 \newlinewhich yields the following table:\newline\begin{tabular}{c|rrcc}\newline\hline \begin{tabular}{c} \newlineCRITICAL POINT \\\newlinex=(x3)3 x=\left(x^{3}\right)^{3} 11\newline\end{tabular} & x=(x3)3 x=\left(x^{3}\right)^{3} 22 & x=(x3)3 x=\left(x^{3}\right)^{3} 33 & x=(x3)3 x=\left(x^{3}\right)^{3} 44 & x=(x3)3 x=\left(x^{3}\right)^{3} 55 \\\newline\hlinex=(x3)3 x=\left(x^{3}\right)^{3} 66 & 00 & 00 & 44 & −16-16 \\\newlinex=(x3)3 x=\left(x^{3}\right)^{3} 77 & −12-12 & −12-12 & 44 & 128128\newline\end{tabular}
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www-awu.aleks.com\newlineVitalSource Bookshelf: Print Reading for Construction\newlineA\newlineALEKS - Zachary Zawacki - Learn\newlinePolynomial and Rational Functions\newlineWriting the equation of a rational function given its graph\newlineZachary\newlineThe figure below shows the graph of a rational function f f .\newlineIt has vertical asymptotes x=−1 x=-1 and x=−5 x=-5 , and horizontal asymptote y=0 y=0 .\newlineThe graph has x x -intercept −4-4 , and it passes through the point (2,2) (2,2) .\newlineThe equation for f(x) f(x) has one of the five forms shown below.\newlineChoose the appropriate form for f(x) f(x) , and then write the equation.\newlineYou can assume that f(x) f(x) is in simplest form.\newlinef(x)=ax−b=□□f(x)=a(x−b)x−c=□(□)□f(x)=a(x−b)(x−c)=□□(□)f(x)=a(x−b)(x−c)(x−d)=□(□)(□)(□)f(x)=a(x−b)(x−c)(x−d)(x−c)=□(□)(□)(□)(□) \begin{array}{ll} f(x)=\frac{a}{x-b} & =\frac{\square}{\square} \\ f(x)=\frac{a(x-b)}{x-c} & =\frac{\square(\square)}{\square} \\ f(x)=\frac{a}{(x-b)(x-c)} & =\frac{\square}{\square(\square)} \\ f(x)=\frac{a(x-b)}{(x-c)(x-d)} & =\frac{\square(\square)}{(\square)(\square)} \\ f(x)=\frac{a(x-b)(x-c)}{(x-d)(x-c)} & =\frac{\square(\square)(\square)}{(\square)(\square)} \end{array} \newlineEspaño
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Let h(x)=1−xtan⁡(x) h(x)=\frac{1-x}{\tan (x)} .\newlineSelect the correct description of the one-sided limits of h h at x=0 x=0 .\newlineChoose 11 answer:\newline(A) lim⁡x→0+h(x)=+∞ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and lim⁡x→0−h(x)=+∞ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(B) lim⁡x→0+h(x)=+∞ \lim _{x \rightarrow 0^{+}} h(x)=+\infty and lim⁡x→0−h(x)=−∞ \lim _{x \rightarrow 0^{-}} h(x)=-\infty \newline(C) lim⁡x→0+h(x)=−∞ \lim _{x \rightarrow 0^{+}} h(x)=-\infty and lim⁡x→0−h(x)=+∞ \lim _{x \rightarrow 0^{-}} h(x)=+\infty \newline(D) lim⁡x→0+h(x)=−∞ \lim _{x \rightarrow 0^{+}} h(x)=-\infty and lim⁡x→0−h(x)=−∞ \lim _{x \rightarrow 0^{-}} h(x)=-\infty
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Let f(x)=x1−cos⁡(x−2) f(x)=\frac{x}{1-\cos (x-2)} .\newlineSelect the correct description of the one-sided limits of f f at x=2 x=2 .\newlineChoose 11 answer:\newline(A) lim⁡x→2+f(x)=+∞ \lim _{x \rightarrow 2^{+}} f(x)=+\infty and lim⁡x→2−f(x)=+∞ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \newline(B) lim⁡x→2+f(x)=+∞ \lim _{x \rightarrow 2^{+}} f(x)=+\infty and lim⁡x→2−f(x)=−∞ \lim _{x \rightarrow 2^{-}} f(x)=-\infty \newline(C) lim⁡x→2+f(x)=−∞ \lim _{x \rightarrow 2^{+}} f(x)=-\infty and lim⁡x→2−f(x)=+∞ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \newline(D) lim⁡x→2+f(x)=−∞ \lim _{x \rightarrow 2^{+}} f(x)=-\infty and lim⁡x→2−f(x)=−∞ \lim _{x \rightarrow 2^{-}} f(x)=-\infty
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Let g(x)=2x+3 g(x)=\frac{2}{x+3} .\newlineSelect the correct description of the one-sided limits of g g at x=−3 x=-3 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→−3+g(x)=+∞ and lim⁡x→−3−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→−3+g(x)=+∞ and lim⁡x→−3−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→−3+g(x)=−∞ and lim⁡x→−3−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→−3+g(x)=−∞ and lim⁡x→−3−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array}
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Let g(x)=2(x−3)2 g(x)=\frac{2}{(x-3)^{2}} .\newlineSelect the correct description of the one-sided limits of g g at x=3 x=3 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→3+g(x)=+∞ and lim⁡x→3−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→3+g(x)=+∞ and lim⁡x→3−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→3+g(x)=−∞ and lim⁡x→3−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→3+g(x)=−∞ and lim⁡x→3−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=-\infty \end{array}
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Let g(x)=−x3sin⁡(x−2) g(x)=-\frac{x}{3 \sin (x-2)} .\newlineSelect the correct description of the one-sided limits of g g at x=2 x=2 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→2+g(x)=+∞ and lim⁡x→2−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→2+g(x)=+∞ and lim⁡x→2−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=-\infty \end{array} \newline(c)\newlinelim⁡x→2+g(x)=−∞ and lim⁡x→2−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→2+g(x)=−∞ and lim⁡x→2−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=-\infty \end{array}
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Let g(x)=1tan⁡2(x) g(x)=\frac{1}{\tan ^{2}(x)} .\newlineSelect the correct description of the one-sided limits of g g at x=0 x=0 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→0+g(x)=+∞ and lim⁡x→0−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→0+g(x)=+∞ and lim⁡x→0−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=-\infty \end{array} \newline(c)\newlinelim⁡x→0+g(x)=−∞ and lim⁡x→0−g(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→0+g(x)=−∞ and lim⁡x→0−g(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=-\infty \end{array}
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Let f(x)=−4x−1 f(x)=-\frac{4}{x-1} .\newlineSelect the correct description of the one-sided limits of f f at x=1 x=1 .\newlineChoose 11 answer:\newline(A) lim⁡x→1+f(x)=+∞ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and lim⁡x→1−f(x)=+∞ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(B) lim⁡x→1+f(x)=+∞ \lim _{x \rightarrow 1^{+}} f(x)=+\infty and lim⁡x→1−f(x)=−∞ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \newline(C) lim⁡x→1+f(x)=−∞ \lim _{x \rightarrow 1^{+}} f(x)=-\infty and lim⁡x→1−f(x)=+∞ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \newline(D) lim⁡x→1+f(x)=−∞ \lim _{x \rightarrow 1^{+}} f(x)=-\infty and lim⁡x→1−f(x)=−∞ \lim _{x \rightarrow 1^{-}} f(x)=-\infty
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Let f(x)=−1(x−1)2 f(x)=-\frac{1}{(x-1)^{2}} .\newlineSelect the correct description of the one-sided limits of f f at x=1 x=1 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→1+f(x)=+∞ and lim⁡x→1−f(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→1+f(x)=+∞ and lim⁡x→1−f(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array} \newline(c)\newlinelim⁡x→1+f(x)=−∞ and lim⁡x→1−f(x)=+∞ \begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→1+f(x)=−∞ and lim⁡x→1−f(x)=−∞ \begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array}
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Let h(x)=−2x(x+1)2 h(x)=-\frac{2 x}{(x+1)^{2}} .\newlineSelect the correct description of the one-sided limits of h h at x=−1 x=-1 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→−1+h(x)=+∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→−1+h(x)=−∞ and lim⁡x→−1−h(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
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Let f(x)=−xln⁡2(x−1) f(x)=\frac{-x}{\ln ^{2}(x-1)} .\newlineSelect the correct description of the one-sided limits of f f at x=2 x=2 .\newlineChoose 11 answer:\newline(A) lim⁡x→2+f(x)=+∞ \lim _{x \rightarrow 2^{+}} f(x)=+\infty and lim⁡x→2−f(x)=+∞ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \newline(B) lim⁡x→2+f(x)=+∞ \lim _{x \rightarrow 2^{+}} f(x)=+\infty and lim⁡x→2−f(x)=−∞ \lim _{x \rightarrow 2^{-}} f(x)=-\infty \newline(C) lim⁡x→2+f(x)=−∞ \lim _{x \rightarrow 2^{+}} f(x)=-\infty and lim⁡x→2−f(x)=+∞ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \newline(D) lim⁡x→2+f(x)=−∞ \lim _{x \rightarrow 2^{+}} f(x)=-\infty and lim⁡x→2−f(x)=−∞ \lim _{x \rightarrow 2^{-}} f(x)=-\infty
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Let f(x)=x5sin⁡(x+1) f(x)=\frac{x}{5 \sin (x+1)} .\newlineSelect the correct description of the one-sided limits of f f at x=−1 x=-1 .\newlineChoose 11 answer:\newline(A)\newlinelim⁡x→−1+f(x)=+∞ and lim⁡x→−1−f(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=+\infty \end{array} \newline(B)\newlinelim⁡x→−1+f(x)=+∞ and lim⁡x→−1−f(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=-\infty \end{array} \newline(C)\newlinelim⁡x→−1+f(x)=−∞ and lim⁡x→−1−f(x)=+∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=+\infty \end{array} \newline(D)\newlinelim⁡x→−1+f(x)=−∞ and lim⁡x→−1−f(x)=−∞ \begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=-\infty \end{array}
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