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

2 Answers
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.
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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 = 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