prove that the value of cosx + secx can never be 3/2.

Answer 1

First make a sketch, here is Wolfram's version
http://www.wolframalpha.com/input/?i=y+%3D+cosx+%2B+1%2Fcosx+for+-%CF%80+to+%2B%CF%80

The vertical lines are asymptotes, and thus will not yield any values. You can see that the min value is
appr 2
Here is a larger picture, showing that it is periodic
http://www.wolframalpha.com/input/?i=y+%3D+cosx+%2B+1%2Fcosx+

let y = cosx + secx
dy/dx = -sinx + secx tanx
= 0 for a max or min

secx tanx = sinx
1/cosx * sinx/cox = sinx
sinx/cos^2 x = sinx
times cos^2 x
sinx = sinx(cos^2 x)
sinx - sinx(cos^2x) = 0
sinx(1 - cos^2 x) = 0
sinx(sin^2 x) = 0
sin^3 x = 0
sinx = 0

sinx x = 0
x = 0, π, 2π, 3π, ...

so let's look at some of the y values
if x = 0,
y = 1 + 1/1 = 2 , ahh, so it was exactly 2
if x = π
y = -1 -1 = -2
So that seems to be our max and min at its corresponding loop
that is, in the main loop between -π and +π, the min is 2

so there are no values of the function that fall between
-2 and +2
and since 3/2 lies in that range, we have proven the case.