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.

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 $\displaystyle \displaystyle \begin{align*} -\sin{x} \end{align*}$...

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.

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.

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?

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 $\displaystyle \cos (-\pi) = \cos(\pi) = -1$.

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?

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.

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:

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