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Consider the function f(x)= (a) Evaluate lim f(x) and X-00 (b) Find the vertical asymptote. For the vertical asymptote x = a, evaluate lim f(x) and lim f(x). x→a x→a* (a) lim f(x) = 00+X lim f(x) = 12 3 2 4x+18x + 18x X→-00 x³+3x² lim f(x), then identify the horizontal asymptotes. X-00

Consider the function f(x)= (a) Evaluate lim f(x) and X-00 (b) Find the vertical asymptote. For the vertical asymptote x = a, evaluate lim f(x) and lim f(x). x→a x→a* (a) lim f(x) = 00+X lim f(x) = 12 3 2 4x+18x + 18x X→-00 x³+3x² lim f(x), then identify the horizontal asymptotes. X-00

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Consider the function** \( f(x) = \frac{4x^3 + 18x^2 + 18x}{x + 3x^2} \). **(a)** Evaluate \( \lim_{{x \to \infty}} f(x) \) and \( \lim_{{x \to -\infty}} f(x) \), then identify the horizontal asymptotes. **(b)** Find the vertical asymptote. For the vertical asymptote \( x = a \), evaluate \( \lim_{{x \to a^-}} f(x) \) and \( \lim_{{x \to a^+}} f(x) \). --- **Solutions:** - **(a)** \(\lim_{{x \to \infty}} f(x) =\) [Blank space for answer] \(\lim_{{x \to -\infty}} f(x) = 12\)
Transcribed Image Text:**Consider the function** \( f(x) = \frac{4x^3 + 18x^2 + 18x}{x + 3x^2} \). **(a)** Evaluate \( \lim_{{x \to \infty}} f(x) \) and \( \lim_{{x \to -\infty}} f(x) \), then identify the horizontal asymptotes. **(b)** Find the vertical asymptote. For the vertical asymptote \( x = a \), evaluate \( \lim_{{x \to a^-}} f(x) \) and \( \lim_{{x \to a^+}} f(x) \). --- **Solutions:** - **(a)** \(\lim_{{x \to \infty}} f(x) =\) [Blank space for answer] \(\lim_{{x \to -\infty}} f(x) = 12\)
Expert Solution
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.

f(x)=\frac{4x^3+18x^2+18x}{x^3+3x^2}

f(x)= \frac{2x\left(2x^2+9x+9\right)}{x^2\left(x+3\right)}

cancel out x

f(x)= \frac{2\left(2x^2+9x+9\right)}{x\left(x+3\right)}

f(x)= \frac{2\left(\left(2x^2+3x\right)+\left(6x+9\right)\right)}{x\left(x+3\right)}

f(x)= \frac{2\left( x\left(2x+3\right)+3\left(2x+3\right)\right)}{x\left(x+3\right)}

f(x)= \frac{2\left( \left(2x+3\right)\left(x+3\right)\right)}{x\left(x+3\right)}

f(x)= \frac{2\left(x+3\right)\left(2x+3\right)}{x\left(x+3\right)}

cancel out x+3

f(x)= \frac{2\left(2x+3\right)}{x}

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