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

Apr 8, 2017

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

Apr 8, 2017

$t = \frac{4}{3} , t = 2$

#### Explanation:

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

$f \left(t\right) = t \sqrt{2 - t}$
$f ' \left(t\right) = \frac{- 3 t + 4}{2 \sqrt{- t + 2}}$

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

$\frac{- 3 t + 4}{2 \sqrt{- t + 2}} = 0 \implies \left(s o l v e\right) \implies t = \frac{4}{3}$

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

$2 \sqrt{- t + 2} = 0 \implies \left(s o l v e\right) \implies t = 2$

Check both $t = 2$ and $t = \frac{4}{3}$ are in $f \left(t\right)$'s domain.

Domain of f(t): $t \le 2$
$\frac{4}{3} \le 2$
$2 \le 2$

Answers: $t = \frac{4}{3} , t = 2$