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

2 Answers
Apr 8, 2017

Answer:

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

Answer:

#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#