How do you use the first and second derivatives to sketch `f(x)= x^4 - 2x^2 +3`?

Given -

`y=x^4-2x^2+3`

`dy/dx=4x^3-4x`

`(d^2y)/(dx^2)=12x^2-4`

To sketch the graph, we have to find for what values of `x` the slope becomes zero. It means at those points the curve turns.

`dy/dx=0 =>4x^3-4x=0`

`4x^3-4x=0`
`4x(x^2-1)=0`
`4x=0`
`x=0`
`x^2-1=0`
`x=+-sqrt1`

`x=1`
`x=-1`

The curve turns when `x=0; x=1; x=-1`

At these points we have to decide whether the curve is concave upwards or concave downwards. For this we need the second derivatives -

At `x=0`

`(d^2y)/(dx^2)=12(0^2)-4=-4<0`

Since the second derivative is less than zero, the curve is concave downwards at `x=0`

The value of the function is -

`y=0^4-2*0^2+3=3`

The curve is concave downwards at `(0, 3)`

At `x=1`

`(d^2y)/(dx^2)=12(1^2)-4=8>0`

Since the second derivative is greater than zero, the curve is concave upwards at `x=0`

The value of the function is -

`y=1^4-2*1^2+3=2`

The curve is concave upwards at `(1, 2)`

At `x=-1`

`(d^2y)/(dx^2)=12(-1^2)-4=8>0`

Since the second derivative is greater than zero, the curve is concave upwards at `x=-1`

The value of the function is -

`y=(-1)^4-2*(-1)^2+3=2`

The curve is concave upwards at `(-1, 2)`