# Suck on maximum and minimum values

• Oct 4th 2012, 04:41 PM
RiderMind
Suck on maximum and minimum values
Hello. I am stuck at a point on this question.

Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].

This is what I have so far (not much I know)

10 cos [-
2π]=10
10 cos [
2π]=10

f(x)=10 sin (x)
0=10 sin (x)
0=sin (x)

Nothing like the examples I was given in class so now I can't figure out what to do. Is anyone able to explain this to me please?

Thanks everybody.
• Oct 4th 2012, 04:54 PM
Prove It
Re: Suck on maximum and minimum values
Quote:

Originally Posted by RiderMind
Hello. I am stuck at a point on this question.

Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].

This is what I have so far (not much I know)

10 cos [-
2π]=10
10 cos [
2π]=10

f(x)=10 sin (x)
0=10 sin (x)
0=sin (x)

Nothing like the examples I was given in class so now I can't figure out what to do. Is anyone able to explain this to me please?

Thanks everybody.

Where did you get the function \displaystyle \displaystyle \begin{align*} f(x) = 10\sin{x} \end{align*} from? If you are trying to differentiate it, first of all, the derivative should be written as \displaystyle \displaystyle \begin{align*} y'(x) \end{align*} or \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} \end{align*}, and second, the derivative of \displaystyle \displaystyle \begin{align*} \cos{x} \end{align*} is \displaystyle \displaystyle \begin{align*} -\sin{x} \end{align*}...
• Oct 4th 2012, 05:00 PM
skeeter
Re: Suck on maximum and minimum values
Quote:

Originally Posted by RiderMind
Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].

you only need a small bit of thought to answer this question (i.e. , no calculation is required).

the maximum value of y is 10 and the minimum value is -10 ... now, think about why.
• Oct 4th 2012, 05:54 PM
richard1234
Re: Suck on maximum and minimum values
Hint: $\displaystyle -1 \le \cos x \le 1$ along the given interval. This alone should immediately lead to the solution.
• Oct 4th 2012, 06:05 PM
RiderMind
Re: Suck on maximum and minimum values
So since -2pi would be the low end and 2pi would be the high end, and plugging those into the equation always equals 10, the minimum is -10 and maximum would be 10? Is that because cos 2pi and -2pi always equals 1?
• Oct 4th 2012, 06:11 PM
richard1234
Re: Suck on maximum and minimum values
Quote:

Originally Posted by RiderMind
So since -2pi would be the low end and 2pi would be the high end, and plugging those into the equation always equals 10, the minimum is -10 and maximum would be 10? Is that because cos 2pi and -2pi always equals 1?

Sure. More specifically, $\displaystyle \cos (-2\pi) = \cos 0 = \cos (2\pi) = 1$, and $\displaystyle \cos (-\pi) = \cos(\pi) = -1$.
• Oct 4th 2012, 06:25 PM
RiderMind
Re: Suck on maximum and minimum values
I see.

The book wants me to explain my answer using transformations. So without using a calculator, could I say that since cos(2pi) is 1, any transformation of interval [-2π, 2π] would yield a stretch of or shrink of 10? Or am I going a little outbound on saying that?
• Oct 5th 2012, 03:18 AM
skeeter
Re: Suck on maximum and minimum values
the transformation was a change in amplitude ... research that term w/r to the graphs of the sine and cosine functions.
• Oct 5th 2012, 04:11 AM
Soroban
Re: Suck on maximum and minimum values
Hello, RiderMind

I'm not sure what "identify" entails . . .

Quote:

$\displaystyle \text{Identify the maximum and minimum values of }f(x) = 10\cos x\text{ in the interval }[\text{-}2\pi,\,2\pi]$

You are expected to know that: $\displaystyle |\cos x| \,\le\,1\,\text{ and }\,\cos(n\pi) \:=\:\pm1$

So we have this list:

. . $\displaystyle \begin{array}{ccc} f(\text{-}2\pi) &=& +10 \\ f(\text{-}\pi) &=& -10 \\ f(0) &=& +10 \\ f(\pi) &=& -10 \\ f(2\pi) &=& +10\end{array}$