# what is the minim and maxim values f or 1+2cos 4x  for all real numbers ?

Differentiating the given function `y = 1 + 2cos 4x` yields:

`(dy)/(dx) = 0 - 8*sin(4x) => (dy)/(dx) = -8*sin(4x)`

You need to solve for x the equation `(dy)/(dx) = 0` to evaluate the critical values of the given function, such that:

`-8*sin(4x) = 0 => sin(4x) = 0 => 4x = (-1)^n sin^(-1) 0 + n*pi`

`4x = 0 + npi => 4x = npi => x = (npi)/4 => {(x = 0),(x = p+-i/4)}`

Replacing 0 for x in equation of the function, yields:

`f(0) = 1 + 2cos4*(0) => f(0) = 1 + 2 = 3`

Replacing `pi/4` for x in equation of the function, yields:

`f(pi/4) = 1 + 2cos4*(pi/4) => f(pi/4) = 1 + 2cos pi = 1 - 2 = -1`

Hence, since the sine function increases over `[0,pi/2]` yields that the function has a minimum value `y = -1` at `x = pi/4` and a maximum value `y = 3,` at `x = 0.`

Approved by eNotes Editorial Team

We know that cosine is always between -1 and +1.
Maximum value of a cosine is +1 and minimum is -1.

So when we consider cos4x maximum would be +1.

And when we consider cos4x minimum would be -1.

`f(x) = 1+2cos4x`

Maximum of `f(x) = 1+2xx1 = 3`

Minimum of `f(x) = 1+2xx(-1) = -1`