How do I find the derivative and critical numbers of x^1/3 (x-4)?

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For this derivative you can either distribute the x^(1/3) OR us the product rule. I'll show you both.

Distribution:

( x^(1/3) )(x - 4) = x^(4/3) - 4x^(1/3)

d/dx [ x^(4/3) - 4x^(1/3) ] = (4/3)x^(1/3) - (4/3)x^(-2/3)


Product Rule:

If f(x) = ( x^(1/3) )(x - 4), g(x) = x^(1/3) and h(x) = x - 4,

then f'(x) = h(x)g'(x) + g(x)h'(x).

g'(x) = (1/3)x^(-2/3) and h'(x) = 1


h(x)g'(x) + g(x)h'(x)

(x - 4)( (1/3)x^(-2/3) ) + (x^(1/3))(1)

(1/3)x^(1/3) - (4/3)x^(-2/3) + x^(1/3)

(4/3)x^(1/3) - (4/3)x^(-2/3)

Which is the same answer we got when we distributed.


When the derivative equals 0 we get the critical numbers.

f(0) does not exist, as 0^(-2/3) = 1/( 0^(2/3) ) = 1/0.

But f(1) is a relative minimum, as (4/3)(1)^(1/3) - (4/3)(1)^(-2/3) = (4/3) - (4/3) = 0.


Hope this helps!...Show more

Well its too easy let the function be y
y=x^1/3 * (x-4)
y=x^4/3-4x^1/3
dy/dx(derivative)=(4/3*x^1/3)-(4/3x^-2/3)
To get the critical points (numbers) just equate the derivative of the function with zero

4/3 (x^1/3-x^-2/3)=0
x^1/3-x^-2/3=0
Take x^1/3 as a common factor
x^1/3 (1-x^-1)=0

x^1/3 =0 therefore x=0
1-x^-1=0 therefore x^-1=1 therefore x=1

Critical numbers are x=0 and x=1...Show more