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! ! !