Question

How do you find local min and max, point of inflection and concavity for the equation `f(x) = 4x^3 + 21x^2 - 294 +7`?

Answer 1

The function has a minimum at `x=0`
The function has a maximum at `x=(-7)/2`
At `x=(-42)/24` there is a point of inflection

Explanation:

Given -
`y=4x^3+21x^2-294+7`
Find the first derivative
`dy/dx=12x^2+42x`
Set it equal to zero. you are doing it to know for which value of 'x' the slope of the curve of the function becomes zero
`dy/dx=6x(2x+7)`
`dy/dx=0 => 6x(2x+7)=0`

`6x = 0`
`x=0`

`2x+7=0`
`2x=-7`
`x=(-7)/2`

Find the second derivative to judge for which value of "x" the function has a minimum, maximum and point of inflection.

`(d^2y)/(dx^2)=24x+42`

At `x=0; (d^2y)/(dx^2)=24(0)+42=42>0`

At `x=0 ; dy/dx=0; (d^2y)/(dx^2)>0`
Hence the function has a minimum at `x=0`

At `x=(-7)/2; (d^2y)/(dx^2)=24((-7)/2)+42= -84+42=-42>0`

At `x=(-7)/2 ; dy/dx=0; (d^2y)/(dx^2)<0`
Hence the function has a maximum at `x=(-7)/2`

To find the point of inflection, set the second derivative equal to zero

`(d^2y)/(dx^2)=0 => 24x+42=0`
`x=(-42)/24`
At `x=(-42)/24` there is a point of inflection

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