range

the range of function
f(x)=tan(2sqrt(x-4)+sqrt(6-x)) is

i know x is (-pi/2,pi/2)

Do you see why the domain is $x - 4 \geq 0$ and $6-x \geq 0$ ?

The range will be between these two points

Quote:

Originally Posted by prat

the range of function
f(x)=tan(2sqrt(x-4)+sqrt(6-x)) is

i know x is (-pi/2,pi/2)

This is quite a subtle problem, in my opinion. You must solve the problem in two fairly complicated overall steps. Let

$f(x)=2\sqrt{x-4}+\sqrt{6-x}.$

You must first find the domain and range of this function. That's step 1. Step 2 is to use the range of this function as the domain of the tangent function, and see if there are any further restrictions. Can you see your way forward?