Where are the minimum and maximum values for f(x) = -2 + 4 cos x on the interval (0,21]?

Step-by-step explanation:

Maximum value is when cos x = 1

So it is -2 + 4(1) = 2.

Minimum value, when  cos x = -1:

= -2  + 4(-1) = -6.

The maximum 2 is reached when x=2pi,4pi, and 6pi.

The minimum -6 is reached when x=pi, 3pi,and 5pi.

Step-by-step explanation:

So let's first look at cos(x) on interval (0,21].

How many rotations is that? Does it at least contain 1 full rotation? If it contains one full rotation that means all the values from -1 to 1 (inclusive) are tagged? If it doesn't contain a full rotation, we might have to dig a little deeper.

So we know x=0 isn't included and that's when cosine is first 1,but this doesn't mean 1 won't be hit later.

Let's figure out the number of rotations:

21/(2pi)=3.3 approximately

This means we make at least 3 rotations.

So this means we definitely will have all the values from -1 to 1 tagged (inclusive).

Now let's look at whole function.

f(x) = -2 + 4 cos x

-2+(-4) to -2+4 will be the range of the function

So the minimum is -6 and the maximum is 2.

So the min occurs when cos(x)=-1 and the max occurs when cos(x)=1.

We have a little over three rotations and remember we can't include x=0.

cos(x)=1

when x=2pi (one full rotation)

when x=4pi (two full rotations)

when x=6pi (three full rotations)

We will stop here because cosine won't be 1 again until a fourth full rotation

cos(x)=-1

when x=pi (half rotation)

When x=3pi (one + half rotation)

When x=5pi (two+half rotation)

We can't include x=7pi (three+half rotation)

because this one is actually not in the interval because 3.5 is more than 3.3 .

The maximum 2 is reached when x=2pi,4pi, and 6pi.

The minimum -6 is reached when x=pi, 3pi,and 5pi.

3x + 18 = 147°

step-by-step explanation:

3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.

i hope i you.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!