Question

What are the critical points of `f(t) = tsqrt(2-t)`?

Answer 1

The critical points are at t=3/4,2

Explanation:

To find the critical points of a function you need to find where the derivative is 0. When the slope is zero there is a horizontal tangent, and thus a maximum or a minimum so a critical feature of the graph.
Derivative:

When finding the derivative do not forget about the product rule and chain rule.

The only time where the derivative is 0 is at t=3/4.

The domain of `sqrt(2-t)` is (- infinity, 2]. Since critical points can occur at endpoints as well this would be considered another critical point.

So the other critical point would be t=2

Answer 2

`t = 4/3 , t = 2`

Explanation:

Critical points (`t = a`) are those points for which `f(t)` is defined and `f'(t)` (its derivative) is `0` or isn't defined.

`f(t) = tsqrt(2-t)`
`f'(t) = (-3t + 4)/(2sqrt(-t+2))`

Let's find where f'(t) = 0:

`(-3t + 4)/(2sqrt(-t+2)) = 0 => (solve) => t = 4/3`

Now where it isn't defined (denominator isn't 0):

`2sqrt(-t+2) = 0 => (solve) => t = 2`

Check both `t=2` and `t = 4/3` are in `f(t)`'s domain.

Domain of f(t): `t <= 2`
`4/3 <= 2` ✓
`2 <= 2` ✓

Answers: `t = 4/3 , t = 2`