fine the coordinates of the turning points of the curve fx = 8x^2-x^4?

Answer 1

y= fx = 8x^2-x^4
dy/dx=16x-4x^3=0,x^2=4
x=+or-2
d^2y/dx^2=16-12x^2
at x=2, d^2y/dx^2=16-48=negative, so x=2gives the maximum value of y
at x=-2, d^2y/dx^2=16-48=negative, so x=-2gives the maximum value of y.
so the turning points are x=2,-2 answer

Answer 2

f(x) = 8x^2 - x^4

Differentiating gives :
f'(x) = 16x - 4x^3

Set f'(x) = 0 for turning points :
16x - 4x^3 = 0

Divide through by 4 :
4x - x^3 = 0

Factorise :
x(4 - x^2) = 0
x(2 - x)(2 + x) = 0

Setting each term to zero gives x = 0 or 2 or -2.

Substituting these values into the original equation gives :
f(x) = 0, 16 and 16, respectively.

Therefore, the turning points are :
(-2, 16), (0, 0) and (2, 16).

Answer 3

Differentiate to get f'(x) = 16x – 4x^3.

Equate to zero; solutions will be the turning points.

16x – 4x^3 = 0
16 – 4x^2 = 0
4x^2 = 16
x^2 = 4
x = +/-2

f(x) = 8 x 4 – 16 = 16 in both cases.

So the turning points are (–2,16) and (2,16).

Answer 4

f `(x) = 16x - 4x³ = 0 for T.P.
4x.(4 - x²) = 0
x = 0, x = ± 2
Turning points are (0,0), (-2,0) , (2,0)