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)).