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

• May 23rd 2010, 07:28 PM
vjelmy
Find the absolute maximum and absolute minimum values of f on the given interval.
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.
• May 23rd 2010, 07:51 PM
Prove It
Quote:

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 $t$.
• May 23rd 2010, 07:55 PM
Prove It
Quote:

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.
• May 23rd 2010, 08:48 PM
vjelmy
im sorry
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
• May 23rd 2010, 09:21 PM
Prove It
Quote:

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.
• May 23rd 2010, 11:03 PM
vjelmy
is a = t?
• May 23rd 2010, 11:07 PM
vjelmy
and how do you get
http://www.mathhelpforum.com/math-he...872e691e-1.gif

http://www.mathhelpforum.com/math-he...71ab0881-1.gif where http://www.mathhelpforum.com/math-he...fc45f909-1.gif
?

the only way i see solving for x is x = arcsin(1/2)
howwwwwwww do you do that?
• May 23rd 2010, 11:23 PM
AllanCuz
Quote:

Originally Posted by vjelmy
and how do you get
http://www.mathhelpforum.com/math-he...872e691e-1.gif

http://www.mathhelpforum.com/math-he...71ab0881-1.gif where http://www.mathhelpforum.com/math-he...fc45f909-1.gif
?

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

Special Triangles
• May 23rd 2010, 11:59 PM
Prove It
Quote:

Originally Posted by vjelmy
and how do you get
http://www.mathhelpforum.com/math-he...872e691e-1.gif

http://www.mathhelpforum.com/math-he...71ab0881-1.gif where http://www.mathhelpforum.com/math-he...fc45f909-1.gif
?

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.