# a curve has equation y=1/12(3x+1)^4,-8x. show that there is a point where x=1/3 and determine whether its a max or a min

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### 1 Answer

To examine the point x= 1/3 on y(x) = (1/12)(3x+1)^4 - 8x.

At x = 1/3, y(1/3)= (1/12)(3/3+1)^4- 8/3= 16/12 - 8/3 = {32-8*32}/12 = -4/3.

Therefore y(1/3) = -4/3.

Therefore y(1/3) = -4/3.

Therefore for x= 1/3 , y = -4/3. Or (x,y) = (1/3, -4/3) is a point on y = (1/12)(3x+1)^4-8x.

To determine whether x= 1/3 a maximum or minimum point on y = (1/2)(3x+1)^4-8x, we proceed as below.

If x = 1/3 is a maximum or a minimum point on y(x) = (1/12)(3x+1)-8x, then y'(1/3) = 0. We examine whether this is true.

y'(x) = dy(x)/dx = (1/12)*4*(3x+1)^3-8 = (1/3)(3x+1)^3-8.

Therefore y'(1/3) = (1/3)(3/3+1)^3 -8 = 8/3 -8 = -16/3.

Or y'(1/3) = -16/3 which is not zero.

So, for the curve y = (1/12)(3x+1)^4 - 8x,

**at the point x= 1/3, y(1/3) = -4/3, there is neither a maximum nor a minimum.**