Finding X & Y Intercept of y= cos(e^(-x) + 2)

[jacy]

Hi,
I have to find x and y intercept of this function

y= cos(e^(-x) + 2)

This is what i have done so far, to find the x intercept i put y= 0

0= cos(e^(-x) + 2)

can i use Cos (A+B)= CosA CosB - SinA SinB here
if i use that then i get

0= cos e^(-x) cos2 - sin e^(-x) sin2

i don't know what to do next. Please help, thanks.

 

[Integral]

Where is the cos function zero?

The argment of the cos must be that which makes cos(a)=0.

 

[jacy]

Where is the cos function zero?

The argment of the cos must be that which makes cos(a)=0.

cos function is zero at 90 degrees. How does that help.

 

[pizzasky]

1) I believe you will need to use radians to solve this problem, because it will not really make sense to carry out algebraic manipulations using an angle in degrees.

2) To find the x-intercept, we need to solve the equation cos(e^{-x} + 2)\ =\ 0. First consider the equation cos(y)\ =\ 0, then see how this can be used to solve the more complicated equation. I don't think the compound angle formula is really needed in this question.

3) Does the question give you a range of x values for which the function is defined? If not, how many x-intercepts can there be? Also, it may be helpful to take note of the range of values e^{-x} can take.

 

[VietDao29]

Have you found out the y-intercept (which is pretty easier) yet?
------------
In most calculus problems, the angle is in radians, so we don't really use degrees here.
\cos x = 0 \Leftrightarrow x = \frac{\pi}{2} + k \pi, \ k \in \mathbb{Z}
Now, just take one minute to look at the unit circle to see if you can get it.
Now if \cos (e ^ {-x} + 2) = 0
What should e^-x + 2 equal?
Can you go from here? :)

 

[jacy]

Thanks pizzasky and VietDao29. No range of x values is given.

(e ^ {-x} + 2) = \frac{\pi}{2}

(e ^ {-x} ) = \frac{\pi}{2} - 2

{-x} = \log \frac{\pi}{2} - 2

So if i substitute the value of -x in the equation i get

cos(pi/2) = 0

 

[HallsofIvy]

However, cosine is periodic. The graph of y= cos(x) crosses the x-axis an infinite number of times and so has an infinite number "x-intercepts".
If no range was given for x, cos(e^{-x}+2)= 0 when e^{-x}+ 2= 2n\pi+ \frac{\pi}{2} or e^{-x}+2= 2n\pi- \frac{\pi}{2} where n can be any integer.

 

[pizzasky]

I do not quite agree with your statement " -x = \log \frac{\pi}{2} - 2". Shouldn't the natural logarithm be used, instead of "log"?

Also, since no range of x values is given, how many x-intercepts can there be? Take a look at VietDao29's post for a hint!

 

[HallsofIvy]

I do not quite agree with your statement " -x = \log \frac{\pi}{2} - 2". Shouldn't the natural logarithm be used, instead of "log"?

Many "advanced" text (beyond calculus) use "log x" to mean the natural logarithm and don't mention the common logarithm.

Also, since no range of x values is given, how many x-intercepts can there be? Take a look at VietDao29's post for a hint!

 

[pizzasky]

Reply

Oops, I did not know that... But shouldn't it be -x = \log (\frac{\pi}{2} - 2) (which, incidentally, is undefined)?

 

[jacy]

Thanks everyone, so there are finite number of x intercepts since no range is given.

 

[pizzasky]

You mean "infinite" number of x intercepts? Whilst there is no limit to the number of x-intercepts, some of the possible intercepts must be rejected. For instance, 1 of the intercepts is calculated to be x = -\log (\frac{\pi}{2} - 2). But is this possible?