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This is one of our many quizzes on End behavior. Click on this to access more such competitive quizzes on Functions.

End behavior Quiz 2

mathematics

 For $f(x)=109{{x}^{121}}$, which of the following graphs represent it in a correct manner?

 Based on the value of a and n for a function $f(x)=a{{x}^{n}}$, we can find the graph of the monomial function for a function $f(x)=a{{x}^{n}}$. 

  $\text{We know that for a function }f(x)=a{{x}^{n}}\text{ where a > 0 and the value of n is odd} \\ \text{Then the graph is as shown below:} \\  $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/100afb8e-c09d-4f70-84ae-d5c78ad2c84e6683084967306230391.png"/>  $ \\ \text{It is clear that the coefficient of }f(x)=109{{x}^{121}}\text{ is greater than 0 and the degree of the monomial is odd} \\ \text{So, the above graph represents the graph of }f(x)=109{{x}^{121}} \\ \text{So, it is clear that the graph of the function }f(x)=109{{x}^{121}}\text{ is graph shown in option C.} $

 Let $f(x)=\dfrac{2{{\left| x \right|}^{3}}+\left| x \right|+2}{{{x}^{3}}+1}$. Which of the following statements are correct?

 The modulus function for x is represented as $\left| x \right|$ where $\left| x \right|=\left\{ \begin{align}  & x,x>0 \\ -x,x<0 \\ \end{align} \right.$. By using the concept of limits, we can find the value of $f(x)=\dfrac{2{{\left| x \right|}^{3}}+\left| x \right|+2}{{{x}^{3}}+1}$ at $x\to -\infty $ and $x\to +\infty $.

  $\text{We have two different cases here to solve.} \\ \text{CASE 1: (Let }x > 0) \\ \Rightarrow f(x)=\dfrac{2{{x}^{3}}+x+2}{{{x}^{3}}+1} \\ \text{As }x\to +\infty  \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{2{{x}^{3}}+x+2}{{{x}^{3}}+1} \right) \\ \text{By using L-Hospital rule, we get} \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{6{{x}^{2}}+1}{3{{x}^{2}}} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( 2+\dfrac{1}{3{{x}^{2}}} \right)=2 \\ \text{As }x\to +\infty,\text{ the value of }f(x)=\dfrac{2{{\left| x \right|}^{3}}+\left| x \right|+2}{{{x}^{3}}+1}\text{ is equal to 2.} \\ \text{CASE 2: (Let }x < 0) \\ \Rightarrow f(x)=\dfrac{-2{{x}^{3}}-x+2}{{{x}^{3}}+1} \\ \text{As }x\to -\infty \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-2{{x}^{3}}-x+2}{{{x}^{3}}+1} \right) \\ \text{By using L-Hospital rule, we get} \\ \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-6{{x}^{2}}-1}{3{{x}^{2}}} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( -2-\dfrac{1}{3{{x}^{2}}} \right)=-2 \\ \text{As }x\to -\infty,\text{ the value of }f(x)=\dfrac{2{{\left| x \right|}^{3}}+\left| x \right|+2}{{{x}^{3}}+1}\text{ is equal to -2.}$

 Find the end behaviour of the function $f(x)={{x}^{6}}+4{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}+1$.

 By observing the values of degree of coefficient and the highest power of the polynomial function, the end behaviour can be determined.

  $\text{In }f(x)={{x}^{6}}+4{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}+1,\text{ the degree of coefficient of }f(x)\text{ is positive and the value of n is equal to n.} \\ \text{So, } f(x)\to +\infty \text{ as x}\to \text{-}\infty \\ f(x)\to +\infty \text{ as x}\to +\infty \\ \text{The graph of the function is shown as below: } \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d6b4fb60-d9d8-4bf0-92be-99b7b7eeeb998050306990611956669.png"/>

 The value of $f(x)=\dfrac{9x+8}{2x+7}$ when $x\to \infty $ is equal to $P$. Then the value of $\left[ P \right]$ is equal to

 $\left[ x \right]$ represents the greater integer function of $x$ where $\left[ x \right]=x-\left\{ x \right\}$. Here $\left\{ x \right\}$ represents the fractional part of $x$. 

  $f(x)=\dfrac{9x+8}{2x+7}=\dfrac{9+\dfrac{8}{x}}{2+\dfrac{7}{x}} \\ \text{As }x\to \infty  \\ f(x)=\dfrac{9+0}{2+0}=\dfrac{9}{2} \\ \Rightarrow \left[ P \right]=\left[ 4.5 \right]=4 \\ \text{So, the value of }\left[ P \right]\text{ is equal to 4.}$

 Which of the following statements is correct based on the end behaviour of the function $f(x)=\dfrac{{{x}^{3}}+3x+1}{9{{x}^{2}}+1}-x-9$? 

 For a function $f(x)$, whose value of the limit is an undetermined form at $x=a$, so to evaluate the limit of $f(x)$ L-Hospital is used.

  $f(x)=\dfrac{{{x}^{3}}+3x+1}{9{{x}^{2}}+1}-x-9=\dfrac{{{x}^{3}}+3x+1-(x+9)\left( 9{{x}^{2}}+1 \right)}{9{{x}^{2}}+1} \\ \Rightarrow f(x)=\dfrac{{{x}^{3}}+3x+1-(9{{x}^{3}}+81{{x}^{2}}+x+9)}{9{{x}^{2}}+1} \\ \Rightarrow f(x)=\dfrac{-8{{x}^{3}}-81{{x}^{2}}+2x-8}{9{{x}^{2}}+1}.....(1) \\ \text{As }x\to +\infty,\text{ the value of }f(x)\text{ is calculated as shown below:} \\ \Rightarrow f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-8{{x}^{3}}-81{{x}^{2}}+2x-8}{9{{x}^{2}}+1} \right)=\dfrac{\infty }{\infty } \\ \text{So, it is clear that the value of }f(x)\text{ represents an indeterminate form. Now we should use the L-Hospital rule.} \\ \Rightarrow f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-8{{x}^{3}}-81{{x}^{2}}+2x-8}{9{{x}^{3}}+1} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-24{{x}^{2}}-162x+2}{27{{x}^{2}}} \right) \\ \Rightarrow f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-48x-162}{54x} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{-48}{54} \right)=\dfrac{-8}{9} \\ \text{So, it is clear that as }x\to +\infty,\text{ the value of }f(x)\text{ is equal to }\dfrac{-8}{9}. \\ \text{As }x\to -\infty,\text{ the value of }f(x)\text{ is equal to }\dfrac{-8}{9}$.