Find the critical numbers of the function : t^4 + t^3 + t^2 + 1

Question

Find the critical numbers of the function : t⁴ + t³ + t² + 1

 

So I found the derivative and set everything = to zero then factored and got this:

 

t(2t - 1)(2t + 2) = 0

 

However, the answer provided in the book is just zero but I also got 1/2 and -1

 

p281, #33

ANSWER: 0

Answer 1

To find the critical numbers, set the derivative of the function equal to zero.  Critical numbers are the location of possible maximum and minimum values.

 

4t³ + 3t² + 2t = 0

 

 

Factor the left side.

 

t(4t² + 3t + 2) = 0

 

 

Attempting to factor completely, we acknowledge that we cannot use FOIL.

 

Set the factors equal to zero.

 

t = 0              and                 4t² + 3t + 2 = 0

      

 

Use the quadratic formula:

 

t = (-b ± √(b² - 4ac))                   for the factor  (4t² + 3t + 2).

 

where:

a = 4

b = 3

c = 2

 

Only real numbers can be critical points.

 

Calculate the discriminant of the quadratic formula to determine if t will be real or complex.

 

b² - 4ac =

 

3² - 4(4)(2) =

 

9 - 32 =

 

-23

 

Since this number is negative, we know that the solution will NOT be a real solution.  That is why t=0 is the only critical point.