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First derivative test Quiz 2

mathematics

 Write the point of inflexion of the function $f\left( b \right)={{b}^{9}}-9b+2003$.

 Use the first derivative test to check for local maxima and local minima for the critical points obtained.

$\text{Consider the function }f\left( b \right)={{b}^{9}}-9b+2003 \\ \text{Differentiate with “b”}. \\ f'\left( b \right)=9{{b}^{8}}-9 \\ \text{Put }\,\,\,\,f'\left( b \right)=0 \\ \Rightarrow 9{{b}^{8}}=9,\Rightarrow b=\pm 1 \\ \text{Make the following tables. For }b=-1 \\ {\,\, } \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9ac37627-86ae-4d73-8155-9765f30572152074564732946131793.png"/> $ \\ $ $b\,=\,-1\text{ is the point of local maxima} \\ b\,=\,1\text{ is the point of local minima. }\\ \text{There is no point of inflexion.}$ 

Write the local minimum value of the function $m\left( s \right)=7{{s}^{6}}-4\,2s+199$

 First find the point of local minima and then substitute this point in the given function. 

 $\text{Consider the function }m\left( s \right) \\ m\left( s \right)=7{{s}^{6}}-42s+199 \\ \text{Differentiate with “s”. } \\ m'\left( s \right)=42{{s}^{5}}-42 \\ \text{Put }m'\left( s \right)\,=\,0 \\ \Rightarrow 42{{s}^{5}}=42 \\ \Rightarrow s=1 \\ \text{Make the following table for s=1}.\\ {\,\,} \\  $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0872ec79-8942-4cb7-92df-c669546f022e5177585845125517183.png"/>$ \\ $ $m'\left( s \right)\text{ changes its sign from negative to positive} \\ \text{So }s\,=\,1\text{ is the point of local minima.} \\ \therefore m\left( 1 \right)=7-42+199=164 \\ \text{Local minimum value is 164.}  $

 Write the number of points of local maxima of the function $a\left( m \right)=17{{m}^{3}}-34{{m}^{2}}+1001$

 Using the first derivative test check local maxima and local minima for the critical points obtained.

 $\text{Consider the function “a(m)”}. \\ a\left( m \right)\,=\,17{{m}^{3}}-34{{m}^{2}}+1001 \\ \Rightarrow \,{{a}^{1}}\,\left( m \right)\,=\,0 \\ \text{Put } {{a}^{1}}\left( m \right)\,=0 \\ 51\,{{m}^{2}}-68m\,=\,0 \\ 17m\left( 3m-4 \right)=0 \\ \Rightarrow m=0,\dfrac{4}{3} \\ \text{Make tables for }m\,=\,0\text{ and }m=\dfrac{4}{3} \\ {\,\, } \\ $.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c5d55cca-5fe8-46f0-9711-7212463062c936546701924591644.png"/>$\\ $ $\text{m=0 is the point of local maxima} \\ m=\dfrac{4}{3}\text{ is the point of local minima.} $

 Let ${{g}_{1}}$ be the critical points of the function $H\left( g \right)=77{{g}^{2}}+308g+123$ and ${{\ell }_{1}}\left( {{\ell }_{1}}>0 \right)$be the critical point of the function $M\left( \ell  \right)=28{{\ell }^{3}}-168\ell +203$. Write the value of ${{\ell }_{1}}\,+\,{{g}_{1}}$.

 Obtain the two critical points by equation ${{H}^{1}}\left( g \right)$ and ${{M}^{1}}\left( \ell  \right)$ to zero and find ${{g}_{1}}$ and ${{\ell }_{1}}$.

 $\text{Consider the function }H\left( g \right) \\ H\left( g \right)=77{{g}^{2}}+308g+123 \\ \Rightarrow {{H}^{1}}\left( g \right)=154g+308 \\ \text{Put }{{H}^{1}}\left( g \right)=0 \\ \Rightarrow 154g+308=0 \\ \Rightarrow g=-2 \\ \text{Consider the function }M\left( \ell  \right) \\ M\left( \ell  \right)=28{{\ell }^{3}}-168\ell +203 \\ {{M}^{1}}\left( \ell  \right)=84{{\ell }^{2}}-168 \\ \text{Put }{{M}^{1}}\left( \ell  \right)=0 \\ \Rightarrow 84{{\ell }^{2}}-168=0 \\ \Rightarrow 84{{\ell }^{2}}=168 \\ \Rightarrow \ell =\pm \sqrt{2} \\ {{g}_{1}}=-2 \text{ and }{{\ell }_{1}}\left( {{\ell }_{1}} > 0 \right)=\sqrt{2} \\ {{g}_{1}}+{{\ell }_{1}}=-2+\sqrt{2}$

 Write the point of local minima of the function $f\left( z \right)={{e}^{3{{z}^{2}}+27z+109}}$.

 Put $f'\left( z \right)$ equal to zero and find the critical point. They use the first derivative test to check whether the critical point is local minima or local maxima.

 $\text{Consider the function }f\left( z \right) \\ f\left( z \right)={{e}^{3{{z}^{2}}+27z+109}} \\ \Rightarrow f'\left( z \right)={{e}^{3{{z}^{2}}+27z+109}}\left( 6z+27 \right) \\ \text{Put }f'\left( z \right)=0 \\ \Rightarrow {{e}^{2{{z}^{2}}+27z+109}}\left( 6z+27 \right)=0 \\ \Rightarrow {{e}^{3{{z}^{2}}+27z+109}}=0 \text{ or } 6z+27=0 \\ \Rightarrow z=-\dfrac{9}{2}\,\,\,\,\,\, [\because {{e}^{z}}=0\text{ for all }Z\in R]\text{ is not possible} \\ \text{Make following table for }z=-\dfrac{9}{2}\\$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ad6e06b1-ce12-49a9-94c2-7d5a1b588f605156591457140373076.png"/>$ \\ f'\left( z \right) \text{ changes sign from negative to positive} \\ \text{So, }z=\dfrac{-9}{2}\text{ is the point of local minima.}$