Locate the absolute extrema of h(t)=t/t-2 on the interval [3,5]

Question

anonymous · Answer

Do you have the derivative of $h(t)$?

anonymous · Answer

yeah if I did it right i got h(t)'=-2/(t-2)^2

anonymous · Answer

Right, so what are the critical points?

anonymous · Answer

I got 0 but I don't know if its right

anonymous · Answer

Critical points are the values of $t$ that make the derivative zero or undefined.

When $t=0$, you have
$$h'(0)=-\frac{2}{(-2)^2}=-\frac{1}{2}$$
so that's not a critical point.

The derivative is zero when the numerator is zero, but that never happens. It's undefined for $t=2$, so you have to test values of $t$ to the left and right of $t=2$.

anonymous · Answer

okay that makes sense thanks for explaining !

myininaya · Answer

well we don't even care about t=2
because we are to look only between 3 and 5 which means you only need to test 3 and 5 to determine what kinda extrema they are (like abs max or min)

anonymous · Answer

No problem. It's possible that the derivative doesn't change from positive to negative or vice versa around $t=2$, which would indicate that no extremum occurs for the critical point. In that case, you would just evaluate the function at the endpoints of the given interval to see what the max/min is.