How do you find the critical numbers of #f(x) = x^4(x-1)^3#?

1 Answer
Aug 20, 2015

Answer:

The critical numbers are #0, 1, "and "4/7#

Explanation:

#f(x) = x^4(x-1)^3#

#f'(x) = 4x^3(x-1)^3 + x^4 3(x-1)^2#

Since #f'# is a polynomial, it is never undefined, so we need only solve:

# 4x^3(x-1)^3 + 3x^4(x-1)^2 = 0#

This is the sum of two terms:

# underbrace(4x^3(x-1)^3) + underbrace(3x^4(x-1)^2) =0#

These terms have common factors of #x^3# and #(x-1)^2#, so remove those:

#x^3(x-1)^2[4(x-1) + 3x] = 0# #" "" "# (simplify in the brackets)

#x^3(x-1)^2[4x-4 + 3x] = 0#

#x^3(x-1)^2(7x-4) = 0#

The roots of the equation and zeros of #f'# are: #0, 1, "and "4/7#

All three of those numbers are in the domain of #f# so they are all critical numbers for #f#.