# [SOLVED] Help! - Values

• Feb 7th 2008, 08:21 AM
weasley74
[SOLVED] Help! - Values
I need help finding out the biggest and smallest values of;

f(x)=2(cosx)^2 - 2sinx + 1

without using the derivative.

Please, need help as soon as possible!
• Feb 7th 2008, 08:25 AM
colby2152
Quote:

Originally Posted by weasley74
I need help finding out the biggest and smallest values of;

f(x)=2(cosx)^2 - 2sinx + 1

without using the derivative.

Please, need help as soon as possible!

You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....
• Feb 7th 2008, 08:37 AM
weasley74
what about the smallest value..? is that just the other way around?
• Feb 7th 2008, 09:03 AM
colby2152
Quote:

Originally Posted by weasley74
what about the smallest value..? is that just the other way around?

Correct!(Yes)
• Feb 7th 2008, 11:40 AM
weasley74
so for minimum x=pi/2 +2npi

and for maximum x=0, x=pi+2npi ?
• Feb 7th 2008, 04:17 PM
Soroban
Hello, weasley74!

Quote:

Find the biggest and smallest values of: . $f(x)\:=\:2\cos^2\!x - 2\sin x + 1$
without using the derivative.

Replace $\cos^2\!x$ with $1 - \sin^2\!x$

We have: . $f(x) \:=\:2(1-\sin^2\!x) - 2\sin x + 1 \quad\Rightarrow\quad f(x) \:=\:-2\sin^2\!x - 2\sin x + 3$

Let $u \,=\,\sin x\!:\;\;f(u) \:=\:-2u^2 - 2u + 3$

This is a down-opening parabola; its maximum is at its vertex.
. . The vertex is at: . $u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}$

Hence, the maximum is: . $f\left(-\frac{1}{2}\right) \:=\:-2\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 3 \;=\;\boxed{\frac{7}{2}}$

Normally, there is no minimum for this parabola.
. . However, the values of $u$ are limited: . $-1 \leq u \leq 1$

The minimum occurs at an endpoint of the interval.

. . And we find that the minimum is: . $f(1) \:=\:\boxed{-1}$

• Feb 7th 2008, 11:24 PM
weasley74
Quote:

Originally Posted by Soroban

This is a down-opening parabola; its maximum is at its vertex.
. . The vertex is at: . $u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}$

how do you get to this point? what is a and b?
• Feb 8th 2008, 02:31 AM
mr fantastic
Quote:

Originally Posted by weasley74
how do you get to this point? what is a and b?

Are you familiar with parabolas? Do you know how to find the turning point of a parabola?

There are many ways to get the turning point - Soroban gave one which clearly you're not familiar with. What about completing the square to get turning point form? You've gotta have met that one ....?

$f(u) = -2u^2 - 2u + 3 = -2 \left( u^2 + u - \frac{3}{2} \right) = -2 \left( \left[ u + \frac{1}{2} \right]^2 - \frac{7}{4} \right) = -2\left(u + \frac{1}{2} \right)^2 + \frac{7}{2}$.

Maximum turning point at $\left(-\frac{1}{2}, \, \frac{7}{2}\right)$, that is, f(u) has a maximum value of $\, \frac{7}{2} \,$ (corresponding to $u = -\frac{1}{2} \Rightarrow \sin x = -\frac{1}{2}$ .....)
• Feb 8th 2008, 02:39 AM
mr fantastic
Quote:

Originally Posted by colby2152
You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....

Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that $\sin x$ can be negative (and hence $- 2 \sin x$ can also make a positive contribution).
• Feb 8th 2008, 04:36 AM
colby2152
Quote:

Originally Posted by mr fantastic
Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that $\sin x$ can be negative (and hence $- 2 \sin x$ can also make a positive contribution).

Yes, that thinking only works best in the 1st quadrant.
• Feb 8th 2008, 05:48 AM
Krizalid
\begin{aligned}
f(u) &= - 2u^2 - 2u + 3\\
&= - \frac{{\left( {4u^2 + 4u - 6} \right)}}
{2}\\
&= - \frac{{(2u + 1)^2 - 7}}
{2}.
\end{aligned}

May be this could be another way to get minimum point.