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Ask QuestionPosted by Ashish Mor 6 years ago
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Posted by Shubham Kumar 6 years ago
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Sia ? 6 years ago
{tex}f'\left( x \right) = - 3{x^3} - 24{x^2} - 45x{/tex}
{tex} = - 3x\left( {{x^2} + 8x + 15} \right) = - 3x\left( {x + 5} \right)\left( {x + 3} \right){/tex}
{tex}f'\left( x \right) = 0 \Rightarrow x = - 5,x = - 3,x = 0{/tex}
{tex}f''\left( x \right) = - 9{x^2} - 48x - 45{/tex}
{tex}= - 3\left( {3{x^2} + 16x + 15} \right){/tex}
{tex}f''\left( 0 \right) = - 45 < 0{/tex}. Therefore, x = 0 is point of local maxima
{tex}f''\left( { - 3} \right) = 18 > 0{/tex}. Therefore, {tex}x = - 3{/tex} is point of local minima
{tex}f''\left( { - 5} \right) = - 30 < 0{/tex}. Therefore, {tex}x = - 5{/tex} is point of local maxima.
Posted by Ashish Mor 6 years ago
- 1 answers
Sia ? 6 years ago
It is clear that gof : R {tex}\rightarrow{/tex} R and fog : R {tex}\rightarrow{/tex} R
Consider {tex}x = \frac{1}{2}{/tex} which lie on (0, # 1)
Now, {tex}(gof)\left( {\frac{1}{2}} \right) = g\left\{ {f\left( {\frac{1}{2}} \right)} \right\} {/tex} = g(1) = [1] = 1
And {tex}(fog)\left( {\frac{1}{2}} \right) = f\left\{ {g\left( {\frac{1}{2}} \right)} \right\} = f\left( {\left[ {\frac{1}{2}} \right]} \right){/tex} = f(0) = 0
{tex}\Rightarrow gof \ne fog{/tex} in (0, 1]
No, fog and gof don't coincide in (0, 1].
Posted by Ashish Mor 6 years ago
- 1 answers
Sia ? 6 years ago
It is clear that gof : R {tex}\rightarrow{/tex} R and fog : R {tex}\rightarrow{/tex} R
Consider {tex}x = \frac{1}{2}{/tex} which lie on (0, # 1)
Now, {tex}(gof)\left( {\frac{1}{2}} \right) = g\left\{ {f\left( {\frac{1}{2}} \right)} \right\} {/tex} = g(1) = [1] = 1
And {tex}(fog)\left( {\frac{1}{2}} \right) = f\left\{ {g\left( {\frac{1}{2}} \right)} \right\} = f\left( {\left[ {\frac{1}{2}} \right]} \right){/tex} = f(0) = 0
{tex}\Rightarrow gof \ne fog{/tex} in (0, 1]
No, fog and gof don't coincide in (0, 1].
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Sia ? 6 years ago
The order is the highest numbered derivative in the equation, while the degree is the highest power to which a derivative is raised. For example: y''+y'=y is a first degree second order differential equation, while (y')^2=y is a second degree first order differential equation.
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