When x = 1,,,, f'( 1) = 2+2f'( 1).

So f'( 1) = -2,,, f'(x) = 2x -4.

So f'(0)=-4.

2.f'(x)=g'(x)+2x

Because the tangent equation of curve g(x) at point (1, g( 1)) is y=2x+ 1.

So g'( 1)=2.

So f' (1) = g' (1)+2x1= 4. That is, =f(x) The slope of the tangent at (1, f( 1)) is _ _ _

3. Solve by derivative method.

If the length is a, the width is s/a.

Circumference y=2(A+S/A)

y ' = 2-2s/a ^ 2 = 0。 When A=sqrt(S) is obtained, y is the smallest.

And ymin=2*2sqrt(S).

4. Let the coordinates of this point be a (x,-x 2+4) and p (0 0,2).

So the slope of AP is (-x 2+4-2)/(x-0) =-x+2/x.

y'=-2x

Because the distance is the shortest and the tangent is perpendicular to the AP line, the slope product is equal to (-1).

So (-2x) (-x+2/x) = 2x 2-4 =- 1.

So x=-√6/2.

So (-√6/2, 5/2)

5。 Tangent = 3x 2 = 3a 2

Let the distance from the focal point of the tangent to the X axis and (a, 0) x = a 3/k = a 3.

S=a/3×a^3× 1/2= 1/6

So a= 1 or-1.

Add, it's hard to write! ! !