What is the f(x)=2cos^2(x) ?

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f(x)=2cos^2(x)

Solution

f (x) = 2 cos^{2} (x)

Solution

Period : π

Domain : - \infty < x < \infty

Range : 0 \leq f (x) \leq 2

X Intercepts : (\frac{π}{2} + πn, 0), Y Intercepts : (0, 2)

Asymptotes : None

Extreme Points : Maximum (πn, 2), Minimum (\frac{π}{2} + πn, 0)

Interval Notation

Domain : (- \infty, \infty)

Range : [0, 2]

Solution steps

Periodicity of

2 cos^{2} (x) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

= \frac{2 π}{2}

Simplify

= π

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

Range of

2 cos^{2} (x) : 0 \leq f (x) \leq 2

Function range definition

Find the minimum and maximum value in each defined interval and unite the results

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

Extreme Points of

2 cos^{2} (x) :

Maximum

(πn, 2),

Minimum

(\frac{π}{2} + πn, 0)

First Derivative Test definition

Find the critical points:

x = πn, x = \frac{π}{2} + πn

2 cos^{2} (x)

Critical point definition

Find where

f^{'} (x)

is equal to zero or undefined

2 cos^{2} (x)

f^{'} (x) = - 2 sin (2 x)

\frac{d}{d x} (2 cos^{2} (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (cos^{2} (x))

Apply the chain rule:

2 cos (x) \frac{d}{d x} (cos (x))

\frac{d}{d x} (cos^{2} (x))

Apply the chain rule:

\frac{df ( u )}{d x} = \frac{df}{d u} \cdot \frac{d u}{d x}

f = u^{2}, u = cos (x)

= \frac{d}{d u} (u^{2}) \frac{d}{d x} (cos (x))

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplify

= 2 u

= 2 u \frac{d}{d x} (cos (x))

Substitute back

u = cos (x)

= 2 cos (x) \frac{d}{d x} (cos (x))

= 2 cos (x) \frac{d}{d x} (cos (x))

\frac{d}{d x} (cos (x)) = - sin (x)

\frac{d}{d x} (cos (x))

Apply the common derivative:

\frac{d}{d x} (cos (x)) = - sin (x)

= - sin (x)

= 2 \cdot 2 cos (x) (- sin (x))

Simplify

2 \cdot 2 cos (x) (- sin (x)) : - 2 sin (2 x)

2 \cdot 2 cos (x) (- sin (x))

Remove parentheses:

(- a) = - a

= - 2 \cdot 2 cos (x) sin (x)

Multiply the numbers:

2 \cdot 2 = 4

= - 4 cos (x) sin (x)

Rewrite using trig identities

- 4 cos (x) sin (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 4 \cdot \frac{sin ( 2 x )}{2}

Simplify

- 4 \cdot \frac{sin ( 2 x )}{2} : - 2 sin (2 x)

- 4 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{sin ( 2 x ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2 sin (2 x)

= - 2 sin (2 x)

= - 2 sin (2 x)

= - 2 sin (2 x)

Solve

- 2 sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

- 2 sin (2 x) = 0

Divide both sides by

- 2

- 2 sin (2 x) = 0

Divide both sides by

- 2

\frac{- 2 sin ( 2 x )}{- 2} = \frac{0}{- 2}

Simplify

sin (2 x) = 0

sin (2 x) = 0

General solutions for

sin (2 x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Identify critical points not in

f (x)

domain

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

All critical points are in domain

x = πn, x = \frac{π}{2} + πn

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

Find intervals:

Decreasing

: πn < x < \frac{π}{2} + πn,

Increasing

: \frac{π}{2} + πn < x < π + πn

Periodicity of

2 cos^{2} (x) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

= \frac{2 π}{2}

Simplify

= π

Combine the critical point(s):

x = πn, x = \frac{π}{2} + πn

with the period:

πn \leq x < π + πn

The function monotone intervals are:

πn < x < \frac{π}{2} + πn, \frac{π}{2} + πn < x < π + πn

Check the sign of

f^{'} (x) = - 2 sin (2 x)

at each monotone interval

Check the sign of

- 2 sin (2 x)

πn < x < \frac{π}{2} + πn :

Negative

Evaluate the derivative at a point on the interval. Take the point

x = 1 + πn

and plug it into

- 2 sin (2 x)

- 2 sin (2 (1 + πn))

Since

f (x)

is periodic, then

f^{'} (x)

is also periodic:

- 2 sin (2 (x + πn)) = - 2 sin (2 x)

- 2 sin (2 \cdot 1)

Simplify

- 2 sin (2)

Refine to a decimal form

- 1.81859 \dots

Negative

Check the sign of

- 2 sin (2 x)

\frac{π}{2} + πn < x < π + πn :

Positive

Evaluate the derivative at a point on the interval. Take the point

x = 2 + πn

and plug it into

- 2 sin (2 x)

- 2 sin (2 (2 + πn))

Since

f (x)

is periodic, then

f^{'} (x)

is also periodic:

- 2 sin (2 (x + πn)) = - 2 sin (2 x)

- 2 sin (2 \cdot 2)

Simplify

- 2 sin (4)

Refine to a decimal form

1.51360 \dots

Positive

Summary of the monotone intervals behavior

Sign Behavior x = πn 0 Maximum πn < x < \frac{π}{2} + πn - Decreasing x = \frac{π}{2} + πn 0 Minimum \frac{π}{2} + πn < x < π + πn + Increasing

Decreasing : πn < x < \frac{π}{2} + πn, Increasing : \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

2 cos^{2} (x) \Rightarrow y = 2

Maximum (πn, 2)

Plug the extreme point

x = \frac{π}{2} + πn

into

2 cos^{2} (x) \Rightarrow y = 0

Minimum (\frac{π}{2} + πn, 0)

Maximum (πn, 2), Minimum (\frac{π}{2} + πn, 0)

Find the range for the interval

- \infty < x < \infty : 0 \leq f (x) \leq 2

Compute the values of the function at the edges of the interval:For

x = πn,

the function value is

f (πn) = 2

x \to π + πn - lim (2 cos^{2} (x)) = 2

x \to π + πn - lim (2 cos^{2} (x))

Plug in the value

x = π + πn

= 2 cos^{2} (π + πn)

Simplify

2 cos^{2} (π + πn) : 2

2 cos^{2} (π + πn)

cos^{2} (π + πn) = 1

cos^{2} (π + πn)

πn = 0

πn

Add the numbers:

0 + 0 = 0

= 0 π

Apply rule

0 \cdot a = 0

= 0

= cos^{2} (π + 0)

π + 0 = π

= cos^{2} (π)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= (- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= 2

The interval has a maximum point at

x = πn + πn

with value

f (πn + πn) = 2

The interval has a minimum point at

x = \frac{π}{2} + πn + πn

with value

f (\frac{π}{2} + πn + πn) = 0

Combine the function value at the edge with the extreme points of the function in the interval:

Minimum function value at the domain interval

- \infty < x < \infty

0

Maximum function value at the domain interval

- \infty < x < \infty

2

Therefore the range of

2 cos^{2} (x)

at the domain interval

- \infty < x < \infty

0 \leq f (x) \leq 2

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (x) \leq 2

Axis interception points of

2 cos^{2} (x) :

X Intercepts

: (\frac{π}{2} + πn, 0),

Y Intercepts

: (0, 2)

x -

axis interception points of

2 cos^{2} (x) : (\frac{π}{2} + πn, 0)

2 cos^{2} (x)

x-axis interception points definition

2 cos^{2} (x) = 0 : x = \frac{π}{2} + πn

2 cos^{2} (x) = 0

Divide both sides by

2

2 cos^{2} (x) = 0

Divide both sides by

2

2 cos^{2} (x) = 0

Divide both sides by

2

\frac{2 cos ^{2} ( x )}{2} = \frac{0}{2}

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Convert solutions to period

π

x = \frac{π}{2} + πn

(\frac{π}{2} + πn, 0)

y -

axis interception point of

2 cos^{2} (x) : (0, 2)

2 cos^{2} (x)

y-axis interception points definition

Solve

y = 2 cos^{2} (0) : 2

2 cos^{2} (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 2 \cdot 1^{2}

Simplify

= 2

(0, 2)

X Intercepts : (\frac{π}{2} + πn, 0), Y Intercepts : (0, 2)

Asymptotes of

2 cos^{2} (x) :

None

Vertical asymptotes of

2 cos^{2} (x) :

None

2 cos^{2} (x)

Go over every undefined point

x = a

and check if at least one of the following statements is true:

x \to a^{-} lim f (x) = \pm \infty

x \to a^{+} lim f (x) = \pm \infty

The function

2 cos^{2} (x)

has no undefined points

No vertical asymptotes

Slant Asymptotes of

2 cos^{2} (x) :

None

2 cos^{2} (x)

Check if at

x \to \pm \infty

the function

y = 2 cos^{2} (x)

behaves as a line,

y = m x + b

where

m \neq = 0

Find an asymptote for

x \to - \infty :

None

Compute

x \to - \infty lim \frac{f ( x )}{x}

to find m:

x \to - \infty lim \frac{f ( x )}{x} = x \to - \infty lim \frac{2 cos ^{2} ( x )}{x} = 0

x \to - \infty lim (\frac{2 cos ^{2} ( x )}{x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 \cdot x \to - \infty lim (\frac{cos ^{2} ( x )}{x})

Apply the Squeeze Theorem:

0

x \to - \infty lim (\frac{cos ^{2} ( x )}{x})

Squeeze Theorem:

x \to - \infty lim (\frac{0}{x}) = 0

x \to - \infty lim (\frac{0}{x})

Apply Infinity Property:

x \to - \infty lim (\frac{c}{x ^{a}}) = 0

= 0

x \to - \infty lim (\frac{1}{x}) = 0

x \to - \infty lim (\frac{1}{x})

Apply Infinity Property:

x \to - \infty lim (\frac{c}{x ^{a}}) = 0

= 0

By the squeeze theorem:

x \to - \infty lim (\frac{cos ^{2} ( x )}{x}) = 0

= 0

= 2 \cdot 0

Simplify

= 0

The slope is zero, therefore

No slant asymptote

Find an asymptote for

x \to \infty :

None

Compute

x \to \infty lim \frac{f ( x )}{x}

to find m:

x \to \infty lim \frac{f ( x )}{x} = x \to \infty lim \frac{2 cos ^{2} ( x )}{x} = 0

x \to \infty lim (\frac{2 cos ^{2} ( x )}{x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 \cdot x \to \infty lim (\frac{cos ^{2} ( x )}{x})

Apply the Squeeze Theorem:

0

x \to \infty lim (\frac{cos ^{2} ( x )}{x})

Squeeze Theorem:

x \to \infty lim (\frac{0}{x}) = 0

x \to \infty lim (\frac{0}{x})

Apply Infinity Property:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 0

x \to \infty lim (\frac{1}{x}) = 0

x \to \infty lim (\frac{1}{x})

Apply Infinity Property:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 0

By the squeeze theorem:

x \to \infty lim (\frac{cos ^{2} ( x )}{x}) = 0

= 0

= 2 \cdot 0

Simplify

= 0

The slope is zero, therefore

No slant asymptote

No slant asymptote

None

Extreme Points of

2 cos^{2} (x) :

Maximum

(πn, 2),

Minimum

(\frac{π}{2} + πn, 0)

First Derivative Test definition

Find the critical points:

x = πn, x = \frac{π}{2} + πn

2 cos^{2} (x)

Critical point definition

Find where

f^{'} (x)

is equal to zero or undefined

2 cos^{2} (x)

f^{'} (x) = - 2 sin (2 x)

\frac{d}{d x} (2 cos^{2} (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (cos^{2} (x))

Apply the chain rule:

2 cos (x) \frac{d}{d x} (cos (x))

\frac{d}{d x} (cos^{2} (x))

Apply the chain rule:

\frac{df ( u )}{d x} = \frac{df}{d u} \cdot \frac{d u}{d x}

f = u^{2}, u = cos (x)

= \frac{d}{d u} (u^{2}) \frac{d}{d x} (cos (x))

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplify

= 2 u

= 2 u \frac{d}{d x} (cos (x))

Substitute back

u = cos (x)

= 2 cos (x) \frac{d}{d x} (cos (x))

= 2 cos (x) \frac{d}{d x} (cos (x))

\frac{d}{d x} (cos (x)) = - sin (x)

\frac{d}{d x} (cos (x))

Apply the common derivative:

\frac{d}{d x} (cos (x)) = - sin (x)

= - sin (x)

= 2 \cdot 2 cos (x) (- sin (x))

Simplify

2 \cdot 2 cos (x) (- sin (x)) : - 2 sin (2 x)

2 \cdot 2 cos (x) (- sin (x))

Remove parentheses:

(- a) = - a

= - 2 \cdot 2 cos (x) sin (x)

Multiply the numbers:

2 \cdot 2 = 4

= - 4 cos (x) sin (x)

Rewrite using trig identities

- 4 cos (x) sin (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 4 \cdot \frac{sin ( 2 x )}{2}

Simplify

- 4 \cdot \frac{sin ( 2 x )}{2} : - 2 sin (2 x)

- 4 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{sin ( 2 x ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2 sin (2 x)

= - 2 sin (2 x)

= - 2 sin (2 x)

= - 2 sin (2 x)

Solve

- 2 sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

- 2 sin (2 x) = 0

Divide both sides by

- 2

- 2 sin (2 x) = 0

Divide both sides by

- 2

\frac{- 2 sin ( 2 x )}{- 2} = \frac{0}{- 2}

Simplify

sin (2 x) = 0

sin (2 x) = 0

General solutions for

sin (2 x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Identify critical points not in

f (x)

domain

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

All critical points are in domain

x = πn, x = \frac{π}{2} + πn

Domain of

2 cos^{2} (x) : - \infty < x < \infty

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

Find intervals:

Decreasing

: πn < x < \frac{π}{2} + πn,

Increasing

: \frac{π}{2} + πn < x < π + πn

Periodicity of

2 cos^{2} (x) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

= \frac{2 π}{2}

Simplify

= π

Combine the critical point(s):

x = πn, x = \frac{π}{2} + πn

with the period:

πn \leq x < π + πn

The function monotone intervals are:

πn < x < \frac{π}{2} + πn, \frac{π}{2} + πn < x < π + πn

Check the sign of

f^{'} (x) = - 2 sin (2 x)

at each monotone interval

Check the sign of

- 2 sin (2 x)

πn < x < \frac{π}{2} + πn :

Negative

Evaluate the derivative at a point on the interval. Take the point

x = 1 + πn

and plug it into

- 2 sin (2 x)

- 2 sin (2 (1 + πn))

Since

f (x)

is periodic, then

f^{'} (x)

is also periodic:

- 2 sin (2 (x + πn)) = - 2 sin (2 x)

- 2 sin (2 \cdot 1)

Simplify

- 2 sin (2)

Refine to a decimal form

- 1.81859 \dots

Negative

Check the sign of

- 2 sin (2 x)

\frac{π}{2} + πn < x < π + πn :

Positive

Evaluate the derivative at a point on the interval. Take the point

x = 2 + πn

and plug it into

- 2 sin (2 x)

- 2 sin (2 (2 + πn))

Since

f (x)

is periodic, then

f^{'} (x)

is also periodic:

- 2 sin (2 (x + πn)) = - 2 sin (2 x)

- 2 sin (2 \cdot 2)

Simplify

- 2 sin (4)

Refine to a decimal form

1.51360 \dots

Positive

Summary of the monotone intervals behavior

Sign Behavior x = πn 0 Maximum πn < x < \frac{π}{2} + πn - Decreasing x = \frac{π}{2} + πn 0 Minimum \frac{π}{2} + πn < x < π + πn + Increasing

Decreasing : πn < x < \frac{π}{2} + πn, Increasing : \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

2 cos^{2} (x) \Rightarrow y = 2

Maximum (πn, 2)

Plug the extreme point

x = \frac{π}{2} + πn

into

2 cos^{2} (x) \Rightarrow y = 0

Minimum (\frac{π}{2} + πn, 0)

Maximum (πn, 2), Minimum (\frac{π}{2} + πn, 0)

Popular Examples

identity csc^2(θ)identity

csc^{2} (θ)

identity 2cos(x)sin(x)identity

2 cos (x) sin (x)

identity 1/(csc(x))identity

\frac{1}{csc ( x )}

identity 1+cot^2(θ)identity

1 + cot^{2} (θ)

identity (cos^2(x))/(sin^2(x))identity

\frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Frequently Asked Questions (FAQ)

What is the f(x)=2cos^2(x) ?
The f(x)=2cos^2(x) is