Question

Find (and classify) the critical points of the following function and determine if they are local max, local min, or neither: f(x) =2x^3 + 3x^2 - 120x

Answer 1

As given by the question

There are given that the function:

`f(x)=2x^3+3x^2-120x`

Now,

To find the critical point, differentiate the given function with respect to x and put the result of function equal to zero

So,

`\begin{gathered} f(x)=2x^3+3x^2-120x \\ f^(\prime)(x)=6x^2+6x-120 \end{gathered}`

Then,

`\begin{gathered} f^(\prime)(x)=0 \\ 6x^2+6x-120=0 \\ x^2+x-20=0 \\ x^2+5x-4x-20=0 \\ x(x+5)-4(x+5) \\ (x-4)(x+5) \\ x=4,\text{ -5} \end{gathered}`

Now,

To find the y-coordinate, we need to substitute the above value, x = 4, -5, into the function f(x)

So,

First put x = 4 into the given function:

`\begin{gathered} f(x)=2x^3+3x^2-120x \\ f(4)=2(4)^3+3(4)^2-120(4) \\ =128+48-480 \\ =-304 \end{gathered}`

And,

Put x = -5 into the function f(x):

`\begin{gathered} f(x)=2x^3+3x^2-120x \\ f(-5)=2(-5)^3+3(-5)^2-120(-5) \\ =-250+75+600 \\ =425 \end{gathered}`

Hence, the critical point is, (4, -304) and (-5, 425).

Now,

To find the local maxima and local minima, we need to find the second derivative of the given function:;

So,

`\begin{gathered} f^(\prime)(x)=6x^2+6x-120 \\ f^(\doubleprime)(x)=12x+6 \end{gathered}`

Now,

The put the value from critical point into the above function to check whether it is maxima or minima.

So,

First put x = 4 into above function:

`\begin{gathered} f^(\doubleprime)(x)=12x+6 \\ f^(\doubleprime)(4)=12(4)+6 \\ f^(\doubleprime)(4)=48+6 \\ f^(\doubleprime)(4)=54 \\ f^(\doubleprime)(4)>0 \end{gathered}`

And,

Put x = -5 into the above function

`\begin{gathered} f^(\doubleprime)(x)=12x+6 \\ f^(\doubleprime)(-5)=12(-5)+6 \\ f^(\doubleprime)(-5)=-60+6 \\ f^(\doubleprime)(-5)=-54 \\ f^(\doubleprime)(-5)<0 \end{gathered}`

Then,

According to the concept, if f''(x)>0 then it is local minima function and if f''(x)<0, then it is local maxima function

Hence, the given function is local maxima at (-5, 425) and the value is -54 and the given function is local minima at point (4, -304) and the value is 54.