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!

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- Feb 7th 2008, 08:21 AMweasley74[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 AMcolby2152
- Feb 7th 2008, 08:37 AMweasley74
what about the smallest value..? is that just the other way around?

- Feb 7th 2008, 09:03 AMcolby2152
- Feb 7th 2008, 11:40 AMweasley74
so for minimum x=pi/2 +2npi

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

Quote:

Find the biggest and smallest values of: .$\displaystyle f(x)\:=\:2\cos^2\!x - 2\sin x + 1$

without using the derivative.

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

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

This is a down-opening parabola; its maximum is at its vertex.

. . The vertex is at: .$\displaystyle u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}$

Hence, the maximum is: .$\displaystyle 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 $\displaystyle u$ are limited: .$\displaystyle -1 \leq u \leq 1$

The minimum occurs at an endpoint of the interval.

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

- Feb 7th 2008, 11:24 PMweasley74
- Feb 8th 2008, 02:31 AMmr fantastic
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 ....?

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

Maximum turning point at $\displaystyle \left(-\frac{1}{2}, \, \frac{7}{2}\right)$, that is, f(u) has a maximum value of $\displaystyle \, \frac{7}{2} \,$ (corresponding to $\displaystyle u = -\frac{1}{2} \Rightarrow \sin x = -\frac{1}{2}$ .....) - Feb 8th 2008, 02:39 AMmr fantastic
- Feb 8th 2008, 04:36 AMcolby2152
- Feb 8th 2008, 05:48 AMKrizalid
$\displaystyle \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.