Calculus help?

Question

1. Given the function v(t) is

(1/2)t when t is greater than zero and less than or equal to four

2 when t is greater than four and less than or equal to six

-t+8 when t is greater than six and less than or equal to eleven

3t-36 when t is greater than eleven and less than or equal to twelve

Find the distance function f(t) that corresponds to v(t).




2. Let f(x) = (2x + 3)^(1/2). Simplify
(f(x + h) - f(x))/h
and evaluate your simplified answer when h = 0.





3. Find (using any method you can think of) cos(x) = x.




Can you give me an answer and maybe an explanation. Thanks so much.

Answer 1

d(t) = t^2 /4for 0 d(t) =4 +2(t -4) 4 d(t) =8-1/2(t^2-36)+(8t-6) 6<=t<=11
d(t) = 11/2+3/2(t^2-11^2)-36(t-11) 11 h can´t be 0 you must
say h==>0
(2x+2h+3)^1/2-(2x+3)^1/2]/h =(2x+2h+3-2x-3)/(2x+2h+3)^1/2+(2x+3)^1/2)h==>2/2(2x+3)^1/2

cos x= x -1<=x<=1
Take y = cosx-x
At the extremes cos (-1) = cos1 =0.5403 and y(-1)>0 and y(1)<0
y´=-senx-1alway<0 so y is decreasing from+ to - and there is only one root.
It is between pi/6 and pi/4