# Thread: find the global max and global min of -3/2arctan(x)-((7x-9)/(2(x^2+1))) on [-1/4,6]

1. ## find the global max and global min of -3/2arctan(x)-((7x-9)/(2(x^2+1))) on [-1/4,6]

I have the problem solved up until I need to use the unit circle to substitute.

-3/2*arctan(-.25) - (-1.75-9)/2.125

From this point, what should I do?

I think after this I would need to substitute 6 into the equation.

Should I take the derivative to find other peaks/dips in the equation?

Thanks
-star

2. Originally Posted by starbless I have the problem solved up until I need to use the unit circle to substitute.

-3/2*arctan(-.25) - (-1.75-9)/2.125

From this point, what should I do?

I think after this I would need to substitute 6 into the equation.

Should I take the derivative to find other peaks/dips in the equation?

Thanks
-star
I don't know what you mean by "I have the problem solved up until I need to use the unit circle to substitute." Asking "should I take the derivatives" makes it look like you haven't actually done anything yet.

Most of the work in this problem is in differentiating to find where any local max or min inside the interval might be. After than you determine the global max and min by evaluating the function itself at those points and at the two endpoints, -1/4 and 6. For that- use a calculator!

#### Search Tags

1 or 4, 3 or 2arctanx7x9 or 2x2, global, max, min 