as level p1 integration 👍👍 the toughest ?

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i.  f'(x) = 2 - 2(x+1)^(-3)

    f''(x) = 6(x+1)^(-4)

   Setting f'(x) = 0 = 2 - 2(x+1)^(-3) = 2[1 - (x+1)^(-3)] 

   For this to work, [1 - 1/(x+1)^3] = 0 which occurs at x=0.

   f''(0) = 6, showing that the curve is concave up at x = 0 and thus a minimum.

2.  Use the distance formula: [(-1/2 - 1)^2 + (3 - 9/4)^2]^(1/2) = (45/16)^1/2

3.  Equation of Line containing the segment above the curve:  y = (-1/2)x + 11/4

     Integrate [(-1/2)x + 11/4] - [2x + (x + 1)^(-2)] over the limits of x = -1/2 to 1.

f(x) = 2x + (x+1)^(-2) =>

f'(x) = 2 - 2(x+1)^(-3) =>

f"(x) = 6(x+1)^(-4).

The point where f'(x) = 0 is where (x+1)^(-3) = 1, which implies x = 0.  At that point, f"(x) > 0, so the point is a minimum value of f.

(ii)  sqrt[(3/2)^2 + (3/4)^2] 

= sqrt(9/4 + 9/16) 

= (3/4)*sqrt(5), around 1.68

(iii) Equation of the line is y = 11/4 - x/2.

Integral from x = -1/2 to x = 9/4 of

[(11/4 - x/2) - (2x + (x+1)^(-2))] dx

= integral from x = -1/2 to +9/4 of

(11/4 - 5x/2 - (x+1)^(-2)) dx.

After integration you have

11x/4 - 5x^2/4 + (x+1)^(-1).

Plug in limits (which is very boring!)...

Question ii) is highschool or even pre high school,math, so I leave you that one.

Question i) is about derivatives, a subject you learned a while back, before integration, so I leave you that one.

Question iii) Imagine thin vertical stripes of height L(x) - f(x) where L(x) is the equation of the line [which you know how to evaluate; high school math again].  Adding (integrating) these  gives:

1

∫  L(x) - ( 2x + (x+1)^(-2) )  dx

-1/2

Done!