# Find all local extrema of the function using the Second Derivative Test where applicable. f(x) =...

## Question:

Find all local extrema of the function using the Second Derivative Test where applicable.

{eq}f(x) = x^{2}\,\left(6 - x\right)^{3} {/eq}

## Local Maxima And Minima:

To evaluate the local extrema of the function i.e. local maxima and local minima of the function our first step is to calculate the critical points and then using the Second Derivative Test we 'll find whether the function at critical point is local maxima or local minima. If

{eq}f''(x)< 0, \text{then x is local maxima}\\ f''(x)> 0, \text{then x is local minima} {/eq}

Given equation is

{eq}f(x) = x^{2}\,\left(6 - x\right)^{3} {/eq}

We have to find all local extrema of the function using the Second Derivative Test where applicable.

First, we find the critical points of the function.

To find the critical points, differentiate the given function with respect to x and then equate it to zero.

{eq}\begin{align} f'(x) &= \frac{\mathrm{d} }{\mathrm{d} x}\left (x^{2}\,\left(6 - x\right)^{3} \right )\\ &=\frac{\mathrm{d} }{\mathrm{d} x}(x^2(216-108x+18x^2-x^3))\\ &=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -x^5+18x^4-108x^3+216x^2 \right )\\ &=-\frac{\mathrm{d} }{\mathrm{d} x}(x^5)+18\frac{\mathrm{d} }{\mathrm{d} x}(x^4)-108\frac{\mathrm{d} }{\mathrm{d} x}(x^3)+216\frac{\mathrm{d} }{\mathrm{d} x}(x^2)\\ &=-5x^4+18(4x^3)-108(3x)+216(2x)\\ &=-5x^4+72x^3-324x^2+432x \end{align} {/eq}

Now, set the derivative equal to zero.

{eq}\begin{align} f'(x) &=0\\ -5x^4+72x^3-324x^2+432x &=0\\ -x\left(5x^3-72x^2+324x-432\right) &=0\\ -x\left(x-6\right)(5x^2-42x+72) &=0\\ -x(x-6)(\left(5x^2-12x\right)+\left(-30x+72\right)) &=0\\ -x(x-6)\left(5x-12\right)\left(x-6\right) &=0\\ -x\left(5x-12\right)\left(x-6\right)^2 &=0\\ x &=0,6, \dfrac{12}{5} \end{align} {/eq}

Now to check whether the point is of local maxima or minima, redifferentiate the above derived function and put the value of critical point.

{eq}\begin{align} f''(x) &=\dfrac{\mathrm{d} }{\mathrm{d} x}(-5x^4+72x^3-324x^2+432x)\\ &=-5(4x^3)+72(3x^2)-324(2x)+432\\ &=-20x^3+216x^2-648x+432 \end{align} {/eq}

At {eq}x=0 {/eq}

{eq}f''(x)\left. \right \}_{x =0} = 432 > 0 {/eq}

As the values comes out to be positive hence the {eq}\displaystyle x = 0 {/eq}, is a point of local minima.

At {eq}x=6 {/eq}

{eq}f''(x)\left. \right \}_{x=6}= 0 {/eq}

As the values comes out to be zero hence the {eq}\displaystyle x = 6 {/eq}, is a point of inflexion.

At {eq}x=\dfrac{12}{5} {/eq}

{eq}\displaystyle f''(x)_{x =\dfrac{12}{5}} = \frac{-3888}{25} {/eq}

As the values comes out to be negative hence the {eq}\displaystyle x = \dfrac{12}{5} {/eq}, is a point of local maxima