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This is one of our many quizzes on Graphing rational functions. Click on this to access more such competitive quizzes on Functions.

Graphing rational functions Quiz 2

mathematics

 Which of the following is correct for the graph of the function $f\left( x \right) = \dfrac{{{x^2} + x + 4}}{{x + 2}}$

 Check for conditions for different types of asymptotes.

$f\left( x \right) = \dfrac{{{x^2} + x + 4}}{{x + 2}} = \left( {x - 1} \right) + \dfrac{6}{{\left( {x + 2} \right)}} \\ $ Thus the slant asymptote is $y = x - 1$

 The inflection point(s) of the function $f\left( x \right) = 4{x^3} + 6{x^2} - 5x$ is

 The inflection points are the point where the concavity of the graph changes.

For inflection points $f''\left( x \right) = 0$, $ \\ \Rightarrow 24x + 12 = 0 \\ \Rightarrow x =  - \dfrac{1}{2} \\  $

 If $f\left( x \right) = \dfrac{{1 - x}}{{1 + x}}$ then which of these is true for $f\left( x \right).f\left( {f\left( x \right)} \right)$

 Simplify the function and use the condition for asymptotes.

$f\left( {f\left( x \right)} \right) = \dfrac{{1 - \dfrac{{1 - x}}{{1 + x}}}}{{1 + \dfrac{{1 - x}}{{1 + x}}}} \\ \Rightarrow \dfrac{{\dfrac{{1 + x - 1 + x}}{{1 + x}}}}{{\dfrac{{1 + x + 1 - x}}{{1 + x}}}} = x \\ f\left( x \right).f\left( {f\left( x \right)} \right) = \dfrac{{x\left( {1 - x} \right)}}{{\left( {1 + x} \right)}} \\ $ Thus the vertical asymptote is $x =  - 1$

 If the vertical and horizontal asymptotes of the function $f\left( x \right) = \dfrac{{x + 4}}{{{x^k} + 3}}$ exists, then

 Check for the conditions of the vertical and the horizontal asymptotes.

For vertical asymptote, $ \\ {x^k} + 3 = 0 \\  \Rightarrow {x^k} =  - 3 \\ $  For $k$ as any even integer we would not get real solutions. Thus $k$ is an odd integer. $ \\ $ For horizontal asymptote, $ \\ y = \mathop {\lim }\limits_{x \to  \pm \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{x + 4}}{{{x^k} + 3}} = \mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{1 + \dfrac{4}{x}}}{{{x^{k - 1}} + \dfrac{3}{x}}} \\ $ For $k < 1$ this limit will not give any finite value. Thus $k \geqslant 1$. $ \\ $ Hence $k$ is a positive odd integer.

 If $f\left( x \right) = \sin x$ and $g\left( x \right) = \cos x$ then

 Use the conditions to check for relations as given in the options. Use a derivative test to check for maximum and minimum.

$f\left( {g\left( x \right)} \right) = \sin \left( {\cos x} \right)$ and $g\left( {f\left( x \right)} \right) = \cos \left( {\sin x} \right) \\ \dfrac{{df\left( {g\left( x \right)} \right)}}{{dx}} =  0  \Rightarrow - \cos \left( {\cos x} \right)\sin x \Rightarrow x = n\pi \;\forall n \in I   \\  $ And $ \\  \dfrac{{dg\left( {f\left( x \right)} \right)}}{{dx}} = 0 \Rightarrow  - \sin \left( {\sin x} \right)\cos x = 0   \\  \Rightarrow x = \dfrac{{n\pi }}{2}\;\forall n \in {I}\;\;\;or\;\;\;x = n\pi \;\forall n \in I   \\ $ Now, $\dfrac{{{d^2}f\left( {g\left( x \right)} \right)}}{{d{x^2}}} =  - \cos \left( {\cos x} \right)\cos x - \sin \left( {\cos x} \right){\sin ^2}x \\ $ At $x = n\pi \;\forall n \in {I_{even}}$, $\dfrac{{{d^2}f\left( {g\left( x \right)} \right)}}{{d{x^2}}} =  - \cos 1 < 0 \\ $  And $\dfrac{{{d^2}g\left( {f\left( x \right)} \right)}}{{d{x^2}}} = \sin \left( {\sin x} \right)\sin x - \cos \left( {\sin x} \right){\cos ^2}x \\ $ At $x = n\pi \;\forall n \in {I_{odd}}$, $\dfrac{{{d^2}f\left( {g\left( x \right)} \right)}}{{d{x^2}}} = \cos 1 > 0 \\ $ Thus, $f\left( {g\left( x \right)} \right)$ attains maximum value at $x = n\pi \;\forall n \in {I_{even}}$ and minimum value at $x = n\pi \;\forall n \in {I_{odd}} \\ $ Similarly, At $x = \dfrac{{n\pi }}{2}\;\forall n \in {I_{even}}$, $\dfrac{{{d^2}g\left( {f\left( x \right)} \right)}}{{d{x^2}}} =  - \cos 0 =  - 1 < 0 \\ $ And at $x = \dfrac{{n\pi }}{2}\;\forall n \in {I_{odd}}$, $\dfrac{{{d^2}g\left( {f\left( x \right)} \right)}}{{d{x^2}}} = \sin 1 > 0 \\ $ Thus, $g\left( {f\left( x \right)} \right)$ attains maximum value at $x = \dfrac{{n\pi }}{2}\;\forall n \in {I_{even}}$ and minimum value at $x = \dfrac{{n\pi }}{2}\;\forall n \in {I_{odd}} \\ \therefore min \left( {f\left( {g\left( x \right)} \right)} \right) < \min \left( {g\left( {f\left( x \right)} \right)} \right)$