Find where values of x have a horizontal tangent line

[KJF]

I know I have to set the derivative of the given function equal to zero. Here is the problem so far:

For what values of x on the interval [0,2pi] does the graph of the following function have a horizontal tangent line?

f(x)= 5+ e^x cos x

...
I did this calculation, hopefully it is correct
f'(x)= (e^x cos x) - (e^x sin x)

My next thought is to set that equal to zero, but I am not sure what to do from there. I blame the fact that I took the pre-requisite courses back in 1998

I thought of factoring out e^x, but know that I can't because the terms are separated by a "-" and not e * or /

 

[stapel]

For what values of x on the interval [0,2pi] does the graph of the following function have a horizontal tangent line?

f(x)= 5+ e^x cos x

...
I did this calculation, hopefully it is correct
f'(x)= (e^x cos x) - (e^x sin x)

My next thought is to set that equal to zero, but I am not sure what to do from there.

Write it out, and see what you can do with the results:

. . . . .\(\displaystyle e^x \cos(x)\, -\, e^x \sin(x)\, =\, 0\)

. . . . .\(\displaystyle e^x\left(\cos(x)\, -\, \sin(x)\right)\, =\, 0\)

Then what? ;-)

 

[KJF]

Write it out, and see what you can do with the results:

. . . . .\(\displaystyle e^x \cos(x)\, -\, e^x \sin(x)\, =\, 0\)

. . . . .\(\displaystyle e^x\left(\cos(x)\, -\, \sin(x)\right)\, =\, 0\)

Then what? ;-)

I am cluless, guess I need the algebra forum
thanks!

 

[HallsofIvy]

One of the things you might learn on the Algebra Forum is the "zero product property"- if p and q are two numbers such that pq= 0, then either p= 0 or q= 0 or both.

Here, the fact that \(\displaystyle e^x(cos(x)- sin(x))= 0\) tells you that either \(\displaystyle e^x= 0\) or \(\displaystyle cos(x)- sin(x)= 0\). You should know (because ln(x) is not defined for x= 0) \(\displaystyle e^x= 0\) has no solution. And, of course, cos(x)- sin(x)= 0 is equivalent to cos(x)= sin(x).