f(t)= 1/t^2+sint+ 1/t+ 1

So f (x) =1/x2+Sint+1/x+1.

2.F (x)' = 3/(5-x) 2+2x/5。 Substitute x=0 and x=2 respectively to get the results.

3. The derivative of parabola at point (2,5) is the slope of tangent, which also passes through this point, so the expression of linear equation is obtained obliquely from this point. y'=2x+ 1=5

Let the tangent y=5x+b, 5=5*2+b, and b=-5, thus the analytical formula is obtained: y=5x-5.

4. If f(x) = x 5-5x+1, it is easy to know that the curve must pass through points (0, 1) and (1, -3), so it is only necessary to prove that f (x) is at (0,1). Let's prove it by derivative.

f '(x)= 5x 4-5 = 5(x ^ 2+ 1)(x ^ 2- 1)。 It is easy to see that the derivative of f' (x) is less than 0 in the range of (0, 1), so it decreases monotonically. So the original function has only one positive real root on (0, 1).

5、

F' (x) = 3x 2+6x-24 = 0, and x=2 or x=-4 is obtained.

Substitute into the extreme value of the original function respectively,

f(x)=x^3+3x^2-24x-20

f(2)=8+ 12-48-20=-40

f(-4)=-64+48+96-20=60

6. The original formula =1/2a * ∫ (1/x+a)-1(x-a) dx =1/2a * ∫ (1/x)

= 1/2a * log(x+a)- 1/2a * log(x-a)

7. Original formula = 2 sinx-cosx+x 2-x+c

8.

d[(x)^( 1/2)]=( 1/2)*dx/[(x)^( 1/2)]

The original formula = 2 ∫ (1, √ 2) (e y) dy = 2 * [e (√ 2)-e]

9. Original formula =∫ (0, √(pi/3)) sin√x d√x

=-cos(√(π/3))+cos(0)=-cos(√(π/3))+ 1