Find the absolute maximum and absolute minimum values of f on the given interval.

**vjelmy** · May 23rd 2010, 06:28 PM

Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(t) = 2 cos t + sin 2t
[0, ( pi)/2]

so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero

Use calculus to find the absolute maximum and minimum values of the function. (Round all answers to three decimal places.)
f(x) = x + 2cos(x)
[0, 2]

again, I know the derivative is 1+-2sinx, but i don't understand how to get a value for x which makes the equation equal to zero.

**Prove It** · May 23rd 2010, 06:51 PM

Originally Posted by vjelmy

Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(t) = 2 cos t + sin 2t
[0, ( pi)/2]

so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero

Actually it's

$f'(t) = -2\sin{t} + 2\cos{2t}$

$= -2\sin{t} + 2(\cos^2{t} - \sin^2{t})$

$= -2\sin{t} + 2(1 - \sin^2{t} - \sin^2{t})$ from the Pythagorean Identity

$= -2\sin{t} + 2(1 - 2\sin^2{t})$

$= -2\sin{t} + 2 - 2\sin^2{t}$

$= -2x^2 - 2x + 2$ , where $x = \sin{t}$ .

This is now a quadratic. So setting the derivative = 0 gives:

$-2x^2 - 2x + 2 = 0$

$x^2 + x - 1 = 0$

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$

$= \frac{-1 \pm \sqrt{5}}{2}$ .

So that means:

$\sin{t} = \frac{-1 + \sqrt{5}}{2}$ or $\sin{t} = \frac{-1 - \sqrt{5}}{2}$ .

Solve these for $t$ .

**Prove It** · May 23rd 2010, 06:55 PM

Originally Posted by vjelmy

Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(t) = 2 cos t + sin 2t
[0, ( pi)/2]

so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero

Use calculus to find the absolute maximum and minimum values of the function. (Round all answers to three decimal places.)
f(x) = x + 2cos(x)
[0, 2]

again, I know the derivative is 1+-2sinx, but i don't understand how to get a value for x which makes the equation equal to zero.

The derivative is not $1 \pm 2\sin{x}$ , it's...

$f'(x) = 1 - 2\sin{x}$ .

If this is 0, then

$0 = 1 - 2\sin{x}$

$2\sin{x} = 1$

$\sin{x} = \frac{1}{2}$

$x = \left\{\frac{\pi}{6}, \pi - \frac{\pi}{6}\right\} + 2\pi n$ where $n \in \mathbf{Z}$

$x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n$ .

Check which ones are in the region $x \in [0, 2]$ ...

Now use the second derivative test to find which of these are maximums and which are minimums.

**vjelmy** · May 23rd 2010, 07:48 PM

but i ammm sooooo confused
the first problem, i don't even understand a little
why is the deriv of sin(2t)) = 2cos2t? the deriv of sinx=cosx. right?

and the second problem, i understand that i can only use pi/6 or 5pi/6
and i know that f" =-2cosx
do i just plug in these values?
I am extremely lost

**Prove It** · May 23rd 2010, 08:21 PM

Originally Posted by vjelmy

but i ammm sooooo confused
the first problem, i don't even understand a little
why is the deriv of sin(2t)) = 2cos2t? the deriv of sinx=cosx. right?

and the second problem, i understand that i can only use pi/6 or 5pi/6
and i know that f" =-2cosx
do i just plug in these values?
I am extremely lost

Using the chain rule:

If $y = \sin{2t}$

Let $u = 2t$ so that $y = \sin{u}$ .

$\frac{du}{dt} = 2$

$\frac{dy}{du} = -\cos{u} = -\cos{2t}$ .

So $\frac{dy}{dt} = \frac{du}{dt}\,\frac{dy}{du}$

$= -2\cos{2t}$ .

For the second problem, the second derivative test states that if $x=a$ is a critical point, then...

If $f''(a) < 0$ then $a$ is a maximum.

If $f''(a) > 0$ then $a$ is a minimum.

If $f''(a) = 0$ then the result is inconclusive.

**vjelmy** · May 23rd 2010, 10:03 PM

is a = t?

**vjelmy** · May 23rd 2010, 10:07 PM

and how do you get

where

?

the only way i see solving for x is x = arcsin(1/2)
howwwwwwww do you do that?

**AllanCuz** · May 23rd 2010, 10:23 PM

Originally Posted by vjelmy

and how do you get

where

?

the only way i see solving for x is x = arcsin(1/2)
howwwwwwww do you do that?

Special Triangles

**Prove It** · May 23rd 2010, 10:59 PM

Originally Posted by vjelmy

and how do you get

where

?

the only way i see solving for x is x = arcsin(1/2)
howwwwwwww do you do that?

In the unit circle, $\sin{\theta}$ is positive in the first and second quadrants, that means that two solutions are

$\arcsin{\frac{1}{2}} = \frac{\pi}{6}$ (Quadrant 1)

and

$\pi - \arcsin{\frac{1}{2}} = \frac{5\pi}{6}$ (Quadrant 2).

But are these the only solutions? Of course not. It depends on how many times you go around the unit circle. So adding an integer multiple of $2\pi$ specifies that you have gone around the unit circle a certain number of times.

Thread: Find the absolute maximum and absolute minimum values of f on the given interval.

LinkBack

Thread Tools

Display

Find the absolute maximum and absolute minimum values of f on the given interval.

im sorry

Similar Math Help Forum Discussions

Find the absolute maximum and absolute minimum values of the function

find absolute minimum and maximum

Find the absolute maximum and absolute minimum

[SOLVED] Determing absolute minimum and maximum values

Absolute maximum & minimum values.

Search Tags