Need help with Stationary Points (Mathematics Differentiation)?

y = (2/3)x^3 + x^2 - 4x

dy/dx = y' = 2x^2 + 2x - 4 = 0

2(x^2 + x - 2) = 0

(x + 2)(x - 1) = 0

x + 2 = 0 or x - 1 = 0

x = -2.............x = 1

y = 20/3.........y = -7/3

d^y/dx^2 = y" = 4x + 2

y"(-2) = -2 < 0

y"(1) = 6 > 0

(-2, 20/3) is a maximum; (1, -7/3) is a minimum.

Your value of 6/3 or 2 is suspect.

Well, dy/dx is the gradient of a line, so the gradient of that line at any point would be dy/dx of that function.

To find the derivative of a function you simply use a formula nx^n-1 = dy/dx of x^n,

So you should apply this 3 . 2/3 . x^2 + 2 . x - 4

Then : 2x^2 + 2x - 4 = dy/dx,

At stationary points the gradient must equal 0, because there is no gradient at the stationary point, think about it, at a flat piece of land is the land rising or lowering, the answer is its not, so dy/dx at stationary points.

so 0 = 2x^2 + 2x - 4, and you know how to factorize, (well you should)

2(x^2 + x -2) (brought 2 outside)

2((x+2)(x-1)) = 0

So the only solutions for this are x = -2 and x = 1,

To find the nature of these stationary points use the second derivative test, so you find the second derivative of the function, just like how you found the derivative of the first function,

So: 2x + 2 = d^2y/dx^2

And when the second derivative is NEGATIVE the stationary point is a MAXIMUM

and vice versa.

So you sub the numbers in; 2(-2) + 2 = -2, therefore it is a max, and 1(2) + 2 = 4, therefore it is a min.

Now you just sub the numbers you obtained to get the y co-ordinate,

2/3 (-2)^3 + (-2)^2 - 4(-2) = 6/3

and

2/3 + 1 - 4 = -7/3

There you go!

y= ⅔x³ + x² -4x

dy/dx = 2x^2 + 2x - 4

dy/dx = 0 at stat points

so 2x^2 + 2x - 4 = 0 at stat points

(2x - 2 )(x + 2) = 0 at stat points

There is a stationary point at x = 1 and x = - 2

When x = 1

y = 2/3 + 1 -4

y = -2 1/3

When x = -2

y = 20/3

To determine their nature we find the second derivative

d^2y/dx^2 = 4x + 2

When x = 1

d^2y/dx^2 = 6

Gradient is increasing so this is a local minimum

When x = -2

d^2y/dx^2 = -6

Gradient is decreasing so this is a local maximum

Yeah looks like your book was wrong on the 6/3 unless I'm wrong of course, but that is impossible.

f ` (x) = 2 x ² + 2x - 4 = 0 for turning points

x ² + x - 2 = 0

(x + 2)(x - 1) = 0

x = - 2 , x = 1

f(-2) = (2/3)(-8) + 4 + 8 = 12 - 16/3 = 20/3

f(1) = 2/3 + 1 - 4 = -7/3

f "(x) = 4x + 2

f "(-2) = - 6

Thus Minimum turning point at x = - 2

f "(1) = 6

Thus Maximum turning point at x = 1

Turning points are:-

(-2 , 20/3 )_____Minimum turning point

( 1 , - 7/3)_____Maximum turning point

you differentiate to get 2x^2 +2x-4 and let that equal zero and solve getting (2x+4)(x-1)=0

2x+4=0 or x-1=0

2x= -4 x=1

x= -2

so x= -2 or 1 and when you put that back in you get y= 6 1/3 -2 1/3

(i assume you meant to put 6 1/3)