min and max

[Tueseve728]

Find the absolute min and max of the function on the given interval.
f(theta)=tan(theta) (-pie/3 >= x>= pie/4)

 

[Unco]

Tan doesn't have any stationary points.

Absolute mins/maxs must occur at the endpoints of your interval.

 

[stapel]

So take the derivative (using the trig-derivative formula they gave you), and set the derivative equal to zero. Solve the resulting trigonometric equation over the given interval.

Where are you stuck?

Eliz.

 

[Tueseve728]

I dont know where to start other than change tan to sec^2

 

[Guest]

different options exist...

f(theta) = tan (theta)

f(x) = sinx / cos x

f (x) = u/v where u = sinx and v = cos x

f ' (x) = ( v du/dx - u dy/dx ) / (v^2)

and so on

 

[stapel]

I dont know where to start other than change tan to sec^2

I'm sorry, but I don't understand what you mean by this. Yes, you could use a trigonometric identity to "change" the tangent:

. . . . .\(\displaystyle \displaystyle \tan{(\theta)} = \sqrt{1 - \sec^2{(\theta)}}\)

But I don't see how this would help. Just differentiate, as suggested. Set the result equal to zero. The square is zero when either of the two (identical) factors is zero, so set a factor equal to zero. Is there any solution?

As suggested earlier, don't forget to check the function's values at the interval endpoints.

Eliz.