
How do you know concavity inflection points and local min/max for
\[f\left( x \right)=2{{x}^{3}}+3{{x}^{2}}-432x\].
Answer
541.5k+ views
Hint: In this problem, we have to determine the local maxima and minima and the inflection points for the given function. We can first find the first derivative and solve the equation using \[f'\left( x \right)=0\] to find the value of x. We can then find the local maxima and minima at x by comparing the values of x. We can then find the second derivative and assume it to 0, to find the x value and if \[f''\left( x \right)>0\] then we can find the concavity inflection points over there.
Complete step by step solution:
We know that the given function is,
\[f\left( x \right)=2{{x}^{3}}+3{{x}^{2}}-432x\]
We can now find the first derivative of the above function,
\[\Rightarrow f'\left( x \right)=6{{x}^{2}}+6x-432\]
We can assume \[f'\left( x \right)=0\],
\[\Rightarrow 6{{x}^{2}}+6x-432=0\]
We can now solve the above equation using the quadratic formula for which a = 6, b = 6, c = -432.
\[\begin{align}
& \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-6\pm \sqrt{36+10368}}{12} \\
& \Rightarrow x=\dfrac{-6\pm 102}{12} \\
& \Rightarrow x=-9,8 \\
\end{align}\]
The local maxima is at x = -9 and the local minima is at x = 8.
Now we can find the inflection points.
We can find the second derivative.
\[\Rightarrow f''\left( x \right)=12x+6\]
We can now assume \[f''\left( x \right)=0\],
\[\begin{align}
& \Rightarrow 12x+6=0 \\
& \Rightarrow x=-\dfrac{1}{2} \\
\end{align}\]
Where, \[f''\left( x \right) > 0,x > -\dfrac{1}{2}\]
We can see that, there is a concavity inflection points,
In \[\left( -\dfrac{1}{2},+\infty \right)\] , the concave is up
In \[\left( -\infty ,-\dfrac{1}{2} \right)\] , the concave is down
Therefore, at \[x=-\dfrac{1}{2}\] there is an inflection point and the local maxima is at x = -9 and the local minima is at x = 8.
Note: Students make mistakes while finding the local maxima and the minima, where the x at its largest number is the local minima and the x at its lowest number is the local maxima. We should remember that at second derivative we have to assume 0, to find the inflection point at x.
Complete step by step solution:
We know that the given function is,
\[f\left( x \right)=2{{x}^{3}}+3{{x}^{2}}-432x\]
We can now find the first derivative of the above function,
\[\Rightarrow f'\left( x \right)=6{{x}^{2}}+6x-432\]
We can assume \[f'\left( x \right)=0\],
\[\Rightarrow 6{{x}^{2}}+6x-432=0\]
We can now solve the above equation using the quadratic formula for which a = 6, b = 6, c = -432.
\[\begin{align}
& \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-6\pm \sqrt{36+10368}}{12} \\
& \Rightarrow x=\dfrac{-6\pm 102}{12} \\
& \Rightarrow x=-9,8 \\
\end{align}\]
The local maxima is at x = -9 and the local minima is at x = 8.
Now we can find the inflection points.
We can find the second derivative.
\[\Rightarrow f''\left( x \right)=12x+6\]
We can now assume \[f''\left( x \right)=0\],
\[\begin{align}
& \Rightarrow 12x+6=0 \\
& \Rightarrow x=-\dfrac{1}{2} \\
\end{align}\]
Where, \[f''\left( x \right) > 0,x > -\dfrac{1}{2}\]
We can see that, there is a concavity inflection points,
In \[\left( -\dfrac{1}{2},+\infty \right)\] , the concave is up
In \[\left( -\infty ,-\dfrac{1}{2} \right)\] , the concave is down
Therefore, at \[x=-\dfrac{1}{2}\] there is an inflection point and the local maxima is at x = -9 and the local minima is at x = 8.
Note: Students make mistakes while finding the local maxima and the minima, where the x at its largest number is the local minima and the x at its lowest number is the local maxima. We should remember that at second derivative we have to assume 0, to find the inflection point at x.
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