Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

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

Answer 1

Luke Alvarez

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

#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#.

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Answer 2

Charles Anctil

To find the critical numbers of ( f(x) = x^4(x-1)^3 ), we first need to find the derivative of the function. Then, we set the derivative equal to zero and solve for ( x ). The critical numbers are the values of ( x ) where the derivative is either zero or undefined.

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

To find the critical numbers, we set the derivative equal to zero and solve for ( x ):

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

Then we solve this equation for ( x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.