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!
Hello, weasley74!
Replace $\displaystyle \cos^2\!x$ with $\displaystyle 1 - \sin^2\!x$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: .$\displaystyle f(1) \:=\:\boxed{-1}$
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 $\displaystyle u = -\frac{1}{2} \Rightarrow \sin x = -\frac{1}{2}$ .....)