Asked by David
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?
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?
Answers
Answered by
Anonymous
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
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
Answered by
David
Thanks!
Answered by
Damon
You are welcome.
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