Question

If f(x)=-8e^x cos(x), find f'(x)
f'(x)= -8 e^x (cos(x)-sin(x))

Find the exact value(s) of x where the function f has a horizontal tangent line on the interval [0,2pi]. If there is more than one, then list them separated by commas.

I don't know what the second part is asking. I already found the derivative. Please help.

Answer 1

f'(x) represent the slope of tangent at the point (x, f(x))
It's asking for horizontal tangent.
@jmacar89 What's the slope of a horizontal line?

Answer 2

@ash2326 that would be 0.

Answer 3

Great so just put f'(x) =0 and find the values of x for which f'(x) is 0
and those values of x which lie in the interval \([0, 2\pi]\) are of our interest:)

Answer 4

Alright, then I would get decimal numbers. Since they give it to me in rad form, how do I change it to rads?

Answer 5

To change degrees into radians, multiply by
\[rads=degrees \times\frac{\pi}{180}\]

Answer 6

@jmacar89 do you get this?

Answer 7

I get the (f,(f(x)) part because it was mentioned earlier in the chapter. I also get that it's looking for the points where it crosses the x-axis between the intervals of [0,2pi] inclusively. I just don't understand how to get the points. I tried substituting the value of 0 for all the x, but I come out with a decimal.

Answer 8

Solve for \(f'(x)=0\)
This means values of x for which f'(x)=0
I'll give you a hint

Answer 9

\[f'(x)= -8 e^x (\cos(x)-\sin x ) \]
f'(x)=0 for horizontal tangents
\[0=-8 e^x (\cos(x)-\sin x ) \]
We know that e^x can't be zero unless x is \(-\infty\) , but we want x to lie between \([0, \pi]\)
so
\[0= \cos x- \sin x \]
@jmacar89 Can you solve for x now?

Answer 10

Honestly, I'm absolutely lost. I got the part where you made f'(x)=0, but I didn't get the rest. so for that equation, am I looking for sin(x)=cos(x)? So both positive or both negative? If so, then what values would they be, and how do I pick them out of the unit circle?

Answer 11

Yeah you are looking for sin x= cos x
So whatever they be positive or negative both should be the same.
\[\cos x= \sin x\]
Let's split \([0, 2\pi]\) into \([0, \pi/2]\)\([\pi/2, \pi]\)\([ \pi, 3\pi/2]\)\([ 3\pi/2, 2\pi]\)
Now let's find the range of our interest
\[\large [0, \pi/2]\Longrightarrow \sin x= +ve,\cos x=+ve\ \bigstar\]
\[\large [\pi/2, \pi]\Longrightarrow \sin x= +ve,\cos x=-ve\ \]
\[\large [\pi, 3\pi/2]\Longrightarrow \sin x= -ve,\cos x=-ve\ \bigstar\]
\[\large [3\pi/2, 2\pi]\Longrightarrow \sin x= -ve,\cos x=+ve\ \]
Star Marked ranges has our answer

Answer 12

@jmacar89 Do you understand this?

Answer 13

Yeah I got the part about how they must both be either positive or negative. How did you get the splitting things up to [0,π/2][π/2,π][π,3π/2][3π/2,2π]?

Answer 14

Any range can be split like this?
\[[0,4]=[0,2][2,4]\]?

Answer 15

Which part did you not understand?

Answer 16

I don't understand how to get the values of the range? I know they limit it to [0,2pi] but that's it. Like for f(x)=-9e^xcosx, the derivative is f'(x)=-9 e^x (cos(x)-sin(x)). I don't know what to do from here. I'm just super confused and forgot all my trig

Answer 17

Do you get the part that we need to have f'(x)=0?

Answer 18

Yes I get that part

Answer 19

You do know that \(e^x\) can't be zero for finite values of x, don't you?

Answer 20

Yup

Answer 21

We have
\[0=-8e^x ( \cos x- \sin x)\]
As e^x is not zero, so
\[0=\cos x-\sin x\]
Any doubts with this?

Answer 22

Nope, I'm with you. so there it would be sinx=cos x

Answer 23

Yeah, now let's solve this

Answer 24

when \(x=\pi/4\)
\[\cos x= \frac{1}{\sqrt 2}\]
and
\[\sin x= \frac{1}{\sqrt 2}\]
We have found one solution, Do you understand this?

Answer 25

Did you just arbitrarily pick a point x? Isn't cos (pi/4)= sqrt(2)/2? and same with sin (pi/4)?

Answer 26

Actually we know that
\[\cos x= \sin x\ for\ x=\pi/4\]
That's why I used it directly and also
\[\large \frac{1}{\sqrt 2}=\frac{1}{\sqrt 2} \times \frac{\sqrt 2}{\sqrt 2}= \frac{\sqrt 2}{2}\]

Answer 27

@jmacar89 did you get my point?

Answer 28

Alright I'm with you

Answer 29

Now we know that in the range \([\pi/2, \pi]\) and \([3\pi/2, 2\pi]\), cos x and sin x have different signs so we can't have any value of x in these ranges satisfy \(\cos x=\sin x\)!!!

Answer 30

Lol alright thanks! I'll try working on this some more and re-read all that.

Answer 31

@jmacar89 do that and if you get stuck anywhere, do let me know. :D:D