This is a calculus homework question but in reality the problem here is understanding the algebra.

Find the Critical Numbers of f(x) = x^(3/5)(4-x)

The product rule gives -x^(3/5) + (12-3x)/(5x^(2/5)

My booklet explains that simplifying this gives you 12-8x / (5x^(2/5).

How did it get to that?

first = x^(3/5)

second = (4-x)

derivative = first derivative of second + second derivative of first

first derivative of second = x^(3/5)*-1

second derivative of first
= (4-x)[ (3/5){x^-(2/5)}
= (1/5)(12 -3x)x^-(2/5)
so
-x^(3/5) + (1/5)(12 -3x)x^-(2/5)

but x^(3/5) = x^(-2/5)x^(5/5)
that is what you missed

Thanks!

You are welcome.

To simplify the expression -x^(3/5) + (12-3x)/(5x^(2/5)), you need to find a common denominator for the two terms.

The first term, -x^(3/5), can be written as -(1/x^(2/5)).

Now, the common denominator for the two terms is x^(2/5).

So, rewriting both terms with the common denominator, you have:

-(1/x^(2/5)) + (12-3x)/(5x^(2/5))

Next, combine the two terms. The numerators of the two terms can be added together because they have a common denominator.

Thus, you have:

(-1 + 12 - 3x)/(x^(2/5))

Simplifying the numerator, you get:

(11 - 3x)/(x^(2/5))

Finally, you can rewrite the expression with the numerator factored out:

(11 - 3x)/(x^(2/5)) = (11 - 3x)/(5x^(2/5))

Therefore, the simplification of -x^(3/5) + (12-3x)/(5x^(2/5)) is (11 - 3x)/(5x^(2/5)).