# The function f(x)=2x^3-6x^2. Find: (a) the minimum and maximum point/-s, intervals of increase and decrease; (b) inflection point/-s, concavity up/down.

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### 1 Answer

`f(x) = 2x^3-6x^2`

To find the extreme points, we should first find the first derivative and then check for the sign of the second derivative to determine whether those points are maxima or minima.

`f'(x) = 6x^2-12x`

For maxima and minima, `f'(x) = 0`

`6x^2-12x = 0`

`x^2-2x = 0`

`x(x-2) = 0`

Therefore there are two points, `x = 0 or x =2.`

`f''(0) = 12x -12`

At `x =0, f''(x) lt 0` , therefore at `x =0` there is a local maximum for f(x) with the value of,

`f(0) = 0 -0 = 0`

**Therefore (0,0) is local maximum point.**

At `x = 2,`

`f''(2) = 12 xx 2 -12`

`f''(2) = 24 -12 = 12 gt 0`

Therefore at x = 2 there is a local minimum with the value of,

`f(2) = 2 xx 2^3 - 6 xx 2^2`

`f(2) = 16 - 24`

`f(2) = -8`

**Therefore (2,-8) is a local minimum for f(x)**

And there are no inflection points in f(x)

In this graph we can identify three main regions.

(i) `-oo to 0`

f(x) is increasing concave down

(ii) `0 to 2`

f(x) is decreasing, concave down

(iii) `2 to +oo`

f(x) is increasing, concave up