The image passes through the point P (0, 1), so e= 1.

At x= 1, the tangent equation is y=x-2, and the tangent point (1,-1).

So f (1) =- 1, f' (1) = 1 (slope of tangent equation).

a+c+ 1=- 1，4a+2c= 1

a=5/2，c=-9/2

fx=2.5x^4-4.5x^2+ 1

Extreme point, f' x = 0, and f'' x has different signs around the extreme point.

f''x=30x^2-9 f'x= 10x^3-9x

According to these two standards, we can find several solutions when f'x=0, substitute the f'x test symbol, and get the extreme point and whether it is the maximum value or the minimum value-this is the standard practice.

However, since fx is already known, it can be clearly seen from a sketch that f0 is the maximum value = 1, the image is a quadratic parabola with an upward opening, the bottom can be reversed, and the two reversal points are symmetrical minima.

When x= plus or minus 9/ 10, there is a minimum value. Do the math yourself.

The combination of numbers and shapes is very important, so we should have a good experience.