Math  /  Calculus

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Local minima x=x= none (b) Find the following left- and right-hand limits at the vertical asymptote x=2x=-2. limx25x2x24=\lim _{x \rightarrow-2^{-}} \frac{5 x^{2}}{x^{2}-4}= \square limx2+5x2x24=\lim _{x \rightarrow-2^{+}} \frac{5 x^{2}}{x^{2}-4}= \square Find the following left- and right-hand limits at the vertical asymptote x=2x=2. limx25x2x24\lim _{x \rightarrow 2^{-}} \frac{5 x^{2}}{x^{2}-4}- ? \square limx2+5x2x24\lim _{x \rightarrow 2^{+}} \frac{5 x^{2}}{x^{2}-4} \square Find the following limits at infinity tg determine any horizontal asymptotes. limx5x2x24=?limx+5x2x24=?\lim _{x \rightarrow-\infty} \frac{5 x^{2}}{x^{2}-4}=? \quad \forall \quad \lim _{x \rightarrow+\infty} \frac{5 x^{2}}{x^{2}-4}=? ? (c) Calculate the second derivative of ff. Find where ff is concave up, concave down, and has inflection points. f(x)=f^{\prime \prime}(x)= \square Union of the intervals where f(x)f(x) is concave up \square Union of the intervals where f(x)f(x) is concave down \square Inflection points x=x= \square Jump to...

Studdy Solution
limx25x2x24=\lim_{x \rightarrow -2^{-}} \frac{5x^2}{x^2 - 4} = -\infty limx2+5x2x24=\lim_{x \rightarrow -2^{+}} \frac{5x^2}{x^2 - 4} = \infty limx25x2x24=\lim_{x \rightarrow 2^{-}} \frac{5x^2}{x^2 - 4} = -\inftylimx2+5x2x24=\lim_{x \rightarrow 2^{+}} \frac{5x^2}{x^2 - 4} = \inftylimx±5x2x24=5\lim_{x \rightarrow \pm\infty} \frac{5x^2}{x^2 - 4} = 5f(x)=40(3x2+4)(x24)3f''(x) = \frac{40(3x^2 + 4)}{(x^2 - 4)^3}Concave up: (,2)(2,)(-\infty, -2) \cup (2, \infty) Concave down: (2,2)(-2, 2) Inflection points: None

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