Question

How do you find and classify local maxima, local minima, and all critical points of `f(x) = x^3+x^2-4x-4`?

Answer 1

There is a local maximum at `(-1.535, ).879` and a local minimum at `(0.869, -6.065)`. The point of inflection is `=(-1/3, -2.593)`.

Explanation:

The function is

`f(x)=x^3+x^2-4x-4`

As this is a polynomial function, the domain is `RR`

Calculate the first derivative

`f'(x)=3x^2+2x-4`

The critical points are when

`f'(x)=0`

That is

`3x^2+2x-4=0`

The solutions to this quadratic equation are

`x=(-2+-sqrt(4+48))/(2*3)=(-2+-sqrt52)/6`

Therefore,

`x_1=(-2-sqrt(52))/6`

`x_2=(-2+sqrt(52))/6`

To determine the nature of thr critical points, you can build either variation chart or calculate the second derivatives.

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `x_1` `color(white)(aaaa)` `x_2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `↗` `color(white)(aaaa)` `↘` `color(white)(aaaa)` `↗`

There is a local maximum at `(-1.535, ).879` and a local minimum at `(0.869, -6.065)`

Calculate the second derivative

`f''(x)=6x+2`

And the point of inflections when

`f''(x)=0`

`=>`, `x=-1/3`

The point of inflection is `=(-1/3, -2.593)`

graph{x^3+x^2-4x-4 [-10, 10, -5, 5]}

Answer 2

Point of maximum: `x=-1.535`

Point of minimum: `x=0.868`

Point of inflection: `x=-1/3`

Explanation:

Given function:

`f(x)=x^3+x^2-4x-4`

`f'(x)=3x^2+2x-4`

`f''(x)=6x+2`

For maximum or minimum points, we must have

`f'(x)=0`

`\therefore 3x^2+2x-4=0`

`x=\frac{-2\pm\sqrt{2^2-4(3)(-4)}}{2(3)}`

`x=\frac{-1\pm\sqrt{13}}{3}`

`=0.868 ,-1.535`

`\implies f''(0.868)=6(0.868)+2=7.208>0`

hence, given function is minimum at `x=0.868`

`\implies f''(-1.535)=6(-1.535)+2=-7.21<0`

hence, given function is maximum at `x=-1.535`

Now, the point of inflection will occur at `f''(x)=0`

`\therefore 6x+2=0`

`x=-1/3`