How do you find the critical numbers for #g(t)=abs(3t-4)# to determine the maximum and minimum?

Question

Scarlett Allison · Accepted Answer

See below.

#g(t) = abs(3t-4) = { (3t-4,"if",t >= 4/3),(-3t+4,"if",t < 4/3 ) :}# #g'(t) = { (3,"if",t >= 4/3),(-3,"if",t < 4/3 ) :}# #g'# is never #0# and is undefined (fals to exist) at #x= 4/3# The only critical number is #4/3#. We see that #g# is decreasing left of #4/3# and increasing on the right. So #g(4/3) = 0# is a local minimum.

Samantha Adkison · Answer

undefined To find the critical numbers for $ g(t) = |3t - 4| $ and determine the maximum and minimum, follow these steps:

1. Find the derivative of $ g(t) $ using the absolute value function:
$$ g'(t) = \frac{d}{dt} |3t - 4| $$

2. Determine where $ g'(t) $ does not exist or is equal to zero. These points are potential critical numbers.

3. Solve for $ t $ when $ g'(t) = 0 $ or when the derivative does not exist.

4. Test the critical numbers in the original function $ g(t) $ to determine whether they correspond to maximum or minimum points.

5. Additionally, check the endpoints of the domain of $ g(t) $, if any, to see if they yield maximum or minimum values.

6. The critical numbers and endpoints with their corresponding function values will give you the maximum and minimum values of $ g(t) $.