what is the minim and maxim values f or 1+2cos 4x for all real numbers ?
3 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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.`
jeew-m | College Teacher | (Level 1) Educator Emeritus
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`
aruv | High School Teacher | (Level 2) Valedictorian
`f(x)=1+2cos(4x)`
`f'(x)=1-8sin(4x)`
`` For maximumor minimum ,f'(x)=0
1-8sin(4x)=0
8sin(4x)=1
sin(4x)=1/8
Solving this equation will give the point of maxima/minima
But if we estimate max /min
`-1<=cos(4x)<=1`
`-2<=cos(4x)<=2`
`1-2<=1+cos(4x)<=1+3`
`max f=3`
`and `
`minf=-1`