find all values in the interval `[-2pi, 2pi]` at which each graph of *f* has a horizontal tangent line. (If someone could just explain the steps on one or two of them I can figure out the rest, I'm just having trouble setting the problem up)

a) `f(x)=4sinx`

b) `f(x)=2x+2cosx`

c) `f(x)=2secx`

d) `f(x)=5tanx`

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You should know that the tangent line to the graph of function is parallel to x axis if derivative of the function, at a point located in `[-2pi,2pi]` , is equal to 0.

Hence, there exists a point `c in [-2pi,2pi]` such that:

`f'(c) = 0`

Considering the function `f(x) = 4 sin x` , first, you need to evaluate its derivative, such that:

`f'(x) = 4 cos x`

You need to replace c for x such that:

`f'(c) = 4 cos c`

You need to solve for `c in [-2pi,2pi]` the equation `f'(c) = 0` , such that:

`4 cos c = 0 => cos c = 0 => c = +-cos^(-1)0 => c = +-pi/2`

**Hence, evaluating the value `c in [-2pi,2pi]` such that the tangent to the graph of the function `f(x) = 4 sin x` , at `x = c` , is parallel to x axis, yields `c = +-pi/2.` **

`f(x)=4sin(x)` and `x in[-2pi,2pi]` .

We want a points on graph of f(x),where tangent will parallel to x axis.

Since slope of x axis is zero. Thus find f'(x) and equate it to zero to get points. Since it is trignometric equation so it has infinite number of points. But for restiction of points interval is given.

Step 1.

`f'(x)=4cos(x)`

`` Step 2.

Let c be the points in `[-2pi,2pi]` such that

`f'(c)=4cos(c)=0`

Now solve the equation

`4cos(c)=0`

`cos(c)=0`

`cos(c0=cos(pi/2)`

`c=2npi+-pi/2`

where n is an integer.

n=0

c=`+-pi/2 in[-2pi,2pi]`

n=1

`c=2pi+-pi/2`

`c=(3pi)/2 in[-2pi,2pi]` but `(5pi)/2!in[-2pi,2pi]`

n=-1

`c=-2pi+-pi/2`

`c=-(3pi)/2 in[-2pi,2pi]` but `-(5pi)/2 !in[-2pi,2pi]`

Thus points are

`+-pi/2,+-(3pi)/2` where tangent will parallel to x axis.

In graph you can see all four points (2 - blue and red) ,(2- green and red)

red line repr. function while blue or green are tangents.

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