Find: For the function below, identify the domain; find the critical points and possible...

Question:

For the function below, identify the domain; find the critical points and possible inflection points; finder intervals on which the function is increasing/decreasing and concave up/down; identify extreme values; and graph the function.

{eq}g(x)=\frac{x^2+7x+12}{(x-3)^2 (x+1)} {/eq}

What I'm needing to do is graph the above function by hand. I need to determine the critical values of the function. I also need to determine the inflection points of the function. After the critical value is found, I need to determine the intervals and the increases and decreases on each interval. Once the inflection points are found, the concavity needs to be determined for each interval created from the inflection points. I need both the x and y values of the absolute extremum. Basically, the problem is requesting the aforementioned information, so that the graph of the function can be accurately produced based on that information.

Studying functions with calculus

A quotient polynomic function with one independent variable has a decreasing interval if its first derivative is negative, and increasing interval if its first derivative is positive, also, a function is concave down if its second derivative is negative.

Answer and Explanation:


The function is:

{eq}\displaystyle \ g(x) = \frac{x^2+7x+12}{(x-3)^2 (x+1)} \\ \displaystyle \ g(x)= {\frac { \left( x+4 \right) \left( x+3 \right) }{ \left( x-3 \right) ^{2} \left( x+1 \right) }} \\ {/eq}


Function domain is all real number except denominator equal zero:


{eq}\displaystyle \mathbb{R}- \{ \; (x-3)^2 (x+1)=0 \; \} \\ \displaystyle (x-3)^2 (x+1)=0 \; \Leftrightarrow \; x= -1 \; \text{ or } \; x= 3 \\ {/eq}

Therefore, function domain is:

{eq}\displaystyle (-\infty,-1)\cup(-1,3)\cup(3,\infty) \\ {/eq}

The first derivative is:

{eq}\displaystyle \ g'(x)={\frac {2\,x+7}{ \left( x-3 \right) ^{2} \left( x+1 \right) }}-2\,{ \frac {{x}^{2}+7\,x+12}{ \left( x-3 \right) ^{3} \left( x+1 \right) }} -{\frac {{x}^{2}+7\,x+12}{ \left( x-3 \right) ^{2} \left( x+1 \right) ^{2}}} \\ \displaystyle \ g'(x)= -{\frac {{x}^{3}+17\,{x}^{2}+49\,x+9}{ \left( x-3 \right) ^{3} \left( x+1 \right) ^{2}}} \\ {/eq}


Critical point at first derivative is zero, so:

{eq}\displaystyle \ g'(x)= 0 \; \Rightarrow \; 0= -{\frac {{x}^{3}+17\,{x}^{2}+49\,x+9}{ \left( x-3 \right) ^{3} \left( x+1 \right) ^{2}}} \; \Leftrightarrow \; x= -13.3910215818105 \; \text{or} \;x= -3.4119994554085 \; \text{or} \;x= -0.196978962781 \\ {/eq}

So, the critical points are:

{eq}\displaystyle (-13.3910215818105,f(-13.3910215818105)) \; \Rightarrow \; (-13.3910215818105 ,- 0.02931243709) \\ \displaystyle ( -3.4119994554085 ,f(-3.4119994554085)) \; \Rightarrow \; (-3.4119994554085 , 0.002442925394 ) \\ \displaystyle ( -0.196978962781 ,f(-0.196978962781)) \; \Rightarrow \; (-0.196978962781 , 1.298818916 ) \\ {/eq}


And first derivative is indeterminate at (denominator equal zero):

{eq}x=-1 \; \text{&} \; x=3 {/eq}

Therefore, increase and decrease interval (first derivative test) :

{eq}\begin{array} Interval & { -\infty < x < -13.39102158 } & { -13.39102158 < x < -3.411999455 } & { -3.411999455 < x < -1 } & { -1 < x < -0.196978963 } & { -0.196978963 < x < 3 } & { 3 < x < \infty } \\ \hline Test \space{} value & { x = -22 } & { x = -10 } & { x = -2 } & { x = -0.5 } & { x = 1.5 } & { x = 3 } \\ Test & { f'x( -22 ) = - < 0 } & { f'x( -10 ) = + > 0 } & { f'x( -2 ) = - < 0 } & { f'x( -0.5 ) = - < 0 } & { f'x( 1.5 ) = + > 0 } & { f'x( 3 ) = - < 0 } \\ Conclusion & {decreasing} & {increasing} & {decreasing} & {decreasing} & { increasing} & {decreasing} \\ \end{array} \\ \therefore \text{ increasing intervals } \; \ ( -13.39102158 , -3.411999455 )\cup( 0 , 1.15 )\cup( -0.196978963 , 3 ) \\ \therefore \text{ decreasing intervals } \; \ ( - \infty , -13.39102158 )\cup( -3.411999455 , -1 )\cup( -1 , -0.196978963 )\cup( 3 , \infty ) \\ {/eq}

Therefore, extrema values at:

{eq}\displaystyle (-13.39102158, f(-13.39102158)) \; \Rightarrow \; (-13.39102158, - 0.02931243709) \; \; \; \text{ Local minimum point} \\ \displaystyle (-3.411999455, f(-3.411999455 )) \; \Rightarrow \; (-3.411999455 , 0.002442925394 ) \; \; \; \text{ Local maximum point} \\ \displaystyle ( -0.196978962781 ,f(-0.196978962781)) \; \Rightarrow \; (-0.196978962781 , 1.298818916 ) \; \; \; \text{ Local minimum point} {/eq}


The second derivative is:

{eq}\displaystyle \ g''(x)= 2\,{\frac {1}{ \left( x-3 \right) ^{2} \left( x+1 \right) }}-4\,{ \frac {2\,x+7}{ \left( x-3 \right) ^{3} \left( x+1 \right) }}-2\,{ \frac {2\,x+7}{ \left( x-3 \right) ^{2} \left( x+1 \right) ^{2}}}+6\,{ \frac {{x}^{2}+7\,x+12}{ \left( x-3 \right) ^{4} \left( x+1 \right) }} +4\,{\frac {{x}^{2}+7\,x+12}{ \left( x-3 \right) ^{3} \left( x+1 \right) ^{2}}}+2\,{\frac {{x}^{2}+7\,x+12}{ \left( x-3 \right) ^{2} \left( x+1 \right) ^{3}}} \\ \displaystyle \ g''(x)= 2\,{\frac {{x}^{4}+27\,{x}^{3}+111\,{x}^{2}+49\,x+60}{ \left( x-3 \right) ^{4} \left( x+1 \right) ^{3}}} \\ {/eq}

Inflection point at second derivative equal zero:

{eq}\displaystyle 0= \ g''(x) \; \Rightarrow \; 0= 2\,{\frac {{x}^{4}+27\,{x}^{3}+111\,{x}^{2}+49\,x+60}{ \left( x-3 \right) ^{4} \left( x+1 \right) ^{3}}} \; \Leftrightarrow \; x= -22.0643175416373 \; \text{&} \; x=-4.608579869143 \\ {/eq}

Then, inflection point at:

{eq}\displaystyle (-22.0643175416373, f(-22.0643175416373)) \; \Rightarrow \; (-22.0643175416373, - 0.02602457590) \\ \displaystyle (-4.608579869143 , f(-4.608579869143 )) \; \Rightarrow \; (-4.608579869143 , - 0.004686155372 ) \\ {/eq}


Concavity intervals:

{eq}\begin{array} Interval & { -\infty < x < -22.06431754 } & { -22.06431754 < x < -4.608579869 } & { -4.608579869 < x < -1 } & { -1 < x < 3 } & { 3 < x < \infty } \\ \hline Test \space{} value & { x = -33 } & { x = -10 } & { x = -2 } & { x = 2 } & { x = 6 } \\ Test & { f''( -33 ) = - < 0 } & { f''( -10 ) = + > 0 } & { f''( -2 ) = - < 0 } & { f''( 2 ) = + > 0 } & { f''( 6 ) = + > 0 } \\ Conclusion & { concave \; downward } & { concave \; upward } & { concave \; downward } & { concave \; upward } & { concave \; upward } \\ \end{array} \\ \therefore \text{ concave upward } \; \ ( -22.06431754 , -4.608579869 )\cup( -1 , 3 )\cup( 3 , \infty ) \\ \therefore \text{ concave downward } \; \ ( - \infty , -22.06431754 )\cup( -4.608579869 , -1 ) \\ {/eq}


Graph:



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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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