Given: f(x)=2+2x2−x4

Calculate the first derivative of the given function.

  f′(x)=4x−4x30=4x−4x31=x2x=0, ±1

The function is increasing in the interval (−∞, −1) .

The function is decreasing in the interval (−1, 0)

The function is increasing in the interval (0, 1) .

The function is decreasing in the interval (1, ∞)

The function is increasing in the interval (−∞, −1) .

The function is decreasing in the interval (−1, 0)

The function is increasing in the interval (0, 1) .

The function is decreasing in the interval (1, ∞)

The value of f′ is changing from positive to negative at x=1 and x=−1

So, the local maxima is (1, f(1))=(1, 3) and (−1, f(−1))=(1, 3)

The value of f′ is changing from negative to positive at x=0 .

So, the local minima is (0, f(0))=(0, 2)

Calculate the second derivative of the given function.

  f″(x)=4−12x24=12x2x=±13

  f″(x)<0 for the interval x<−13 . So, the graph is concave down for (−∞,−13) .

  f″(x)>0 for the interval −13<x<13 . So, the graph is concave up for (−13, 13) .

  f″(x)<0 for the interval x>13 . So, the graph is concave down for (13, ∞) .

The inflection point is (13, 239) and (−13, 239) .

Given: f(x)=2+2x2−x4

The graph of the result obtained is shown in figure below.

Figure (1)

Therefore, the required graph is shown in figure (1).