Science:Math Exam Resources/Courses/MATH102/December 2015/Question 09

MATH102 December 2015

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Question 09
Match the function ((a), (b), (c)) to its sketch (one of (D), (E), (F), (G)). (a) ${\frac {3x^{2}+x^{3}}{3+2x^{3}}}$ , (b) ${\frac {x^{3}}{3+2x^{3}}}$ (c) ${\frac {x^{3}+x^{4}}{1+2x^{5}}}$

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Hint
For each given function, consider the function value at $x=0$ , the limit at infinity $\infty$ , and the sign of the function on the interval $-1<x<0$ .

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (a) Let $f(x)={\frac {3x^{2}+x^{3}}{3+2x^{3}}}$ . Since $f(0)={\frac {3\cdot 0^{2}+0^{3}}{3+2\cdot {0}^{3}}}=0$ , the graph (G) is out. Considering $\lim _{x\to \infty }f(x)=\lim _{x\to \infty }{\frac {3x^{2}+x^{3}}{3+2x^{3}}}=\lim _{x\to \infty }{\frac {\frac {3x^{2}+x^{3}}{x^{3}}}{\frac {3+2x^{3}}{x^{3}}}}=\lim _{x\to \infty }{\frac {{\frac {3}{x}}+1}{{\frac {3}{x^{3}}}+2}}=2$ , the graph (E) is also out. Finally, we can factor out the numerator and rewrite $f$ as $f(x)={\frac {3x^{2}+x^{3}}{3+2x^{3}}}=x^{2}\cdot {\frac {3+x}{3+2x^{3}}}$ . Then, for $-1<x<0$ , the denominator of the fraction is positive $3+2x^{3}>3-2>0$ , and so is the numerator $3+x>3-1>2$ . It follows that for $-1<x<0$ , $f(x)>0$ , and hence (F) is out. To sum, it matches with (D). (b) Let $g(x)={\frac {x^{3}}{3+2x^{3}}}$ . Using $g(0)=0$ and $\lim _{x\to \infty }g(x)=\lim _{x\to \infty }{\frac {x^{3}}{3+2x^{3}}}=\lim _{x\to \infty }{\frac {1}{{\frac {3}{x^{3}}}+2}}={\frac {1}{2}}$ , as in the solution for (a), the graphs (G) and (E) are out. But for $-1<x<0$ , we have $-1<x^{3}<0$ and $3+2x^{3}>0$ , which implies that $f(x)<0$ . i.e., (D) is out. Therefore, it matches with (F). (c) Let $h(x)={\frac {x^{3}+x^{4}}{1+2x^{5}}}$ We can easily see that $h(0))=0$ and $\lim _{x\to \infty }h(x)=\lim _{x\to \infty }{\frac {\frac {x^{3}+x^{4}}{x^{5}}}{\frac {1+2x^{5}}{x^{5}}}}=\lim _{x\to \infty }{\frac {{\frac {1}{x^{2}}}+{\frac {1}{x}}}{{\frac {1}{x^{5}}}+2}}=0$ . Then, (D), (F), and (G) are out, so that it matches with (E) Answer $\color {blue}(a)-(D),(b)-(F),(c)-(E)$