Question

calculus help
http://prntscr.com/deac3s

Answer 1

i found c to be 1/e, can you check to see if that is right

Answer 2

\[\large\rm y'-y \sin x=0\]Integrating factor: \(\large\rm \mu=e^{\cos x}\)
(If you did separate of variables instead, that's fine also).
Multiplying through by our integrating factor, and doing some other stuff,\[\large\rm \int\limits\left(y e^{\cos x}\right)'=\int\limits 0\]giving us,\[\large\rm y e^{\cos x}=c\qquad\to\qquad y(x)=ce^{-\cos x}\]Plugging in our data to find c,
ya ya ya c=1/e.

Answer 3

So your function is,\[\large\rm y(x)=e^{-1-\cos x}\]and they want to know the range of this function.
Remember any particular things about the exponential function? :)

Answer 4

um it like doesn't reach a certain number depending on what your function is

Answer 5

Oh, actually ...
for this problem... it might be easier to start with the range of cosine,\[\large\rm -1\le \cos x\le 1\]And build our function in the middle.

Answer 6

We can multiply through by -1,
Everything switches sign, even the inequalities flip,\[\large\rm 1\ge -\cos x\ge -1\]Let's turn it around, it's hard to read with the inequalities going this way,\[\large\rm -1\le -\cos x\le 1\]Oh -cos x has the same range as cos x, I guess I was being silly on that first step hehe.

Anyway, subtract 1 from all sides,\[\large\rm -2\le -1-\cos x\le 0\]And then exponentiate each side,\[\large\rm e^{-2}\le e^{-1-\cos x}\le e^0\]

Answer 7

What do you think?
Is that super confusing? :o

Answer 8

that's the range? How come it's not just [-1,1]?

Answer 9

I thought the function is y=e^-cosx

Answer 10

Can u cancel your question now?

Answer 11

The function is:
\(\large\rm y=Ce^{-cos x}\)
if you plug in your C you get,
\(\large\rm y=\frac1 e e^{-cos x}\)

I applied an exponent rule to bring the 1/e into the other e,
\(\large\rm y=e^{-1}e^{-cos x}=e^{-1-cos x}\)

Answer 12

oh ok, i see

Answer 13

so what you had above is the range?

Answer 14

We started with the range of -cos x, which is [-1,1].
And we applied some algebra steps to "build" our function in the middle.

It looks like it changed the boundaries to [e^{-2}, e^0] right?
Which is something like [0.135, 1].

Answer 15

ok that makes sense. Thank you so much for the help!

Answer 16

np :)