# anyone know how to solve this differentiation？ One of this project is to design GSP project. You are given a task involving graph polynomials. `f (x) = 2x^6 +3x*(3x^5 + x^ 3) - 2x^2` . Your...

## anyone know how to solve this differentiation？

One of this project is to design GSP project. You are given a task involving graph polynomials. `f (x) = 2x^6 +3x*(3x^5 + x^ 3) - 2x^2` . Your graph should include the following:

Use the graph of f' to estimate the maximum and minimum points.

Hard copy the graph to be submitted.

Calculus involved in determining the maximum and minimum points to be discussed.

*print*Print*list*Cite

### 1 Answer

You should use the first derivative to find the critical points, hence, you need to solve the equation `f'(x) = 0` such that:

`f'(x) = 2*6x^5 + 9*6x^5 + 3*4x^3 - 2*2x`

`f'(x) = 66x^5 + 12x^3 - 4x`

You need to solve the equation `f'(x) = 0` such that:

`66x^5 + 12x^3 - 4x = 0`

Factoring out 2x yields:

`2x(33x^4 + 6x^2 - 2) = 0`

`2x = 0 => x = 0`

You need to solve the bi-quadratic equation `33x^4 + 6x^2 - 2 = 0` , hence, you should come up with the substitution x^2 = y such that:

`33y^2 + 6y - 2 = 0 `

You need to use quadratic formula such that:

`y_(1,2) = (-6+-sqrt(36 + 264))/66`

`y_(1,2) = (-6+-sqrt300)/66`

`y_(1,2) = (-6+-10sqrt3)/66`

`y_(1,2) = (-3+-5sqrt3)/33`

You need to solve for x the equations `x^2 = y_(1,2)` such that:

`x^2 = (-3+5sqrt3)/33 => x_(1,2) = +-sqrt((-3+5sqrt3)/33)`

Since `y_2 < 0,` the equation `x^2 = y_2` cannot be solved for real x.

Hence, evaluating the critical points yields `x = -sqrt((-3+5sqrt3)/33) , x = 0` and `x = sqrt((-3+5sqrt3)/33).`

**Notice that the graph shows you that the function reaches the maximum at `x = 0` and it reaches the minimums at `x_(1,2) = +-sqrt((-3+5sqrt3)/33).` **