What are the critical points of #g(x)=x^(2/5)(x-10)#?

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Answer 1 · 2015-10-29T03:34:06

#0# and #20/7#

#g(x)=x^(2/5)(x-10)#

Note first that #"Dom(g) = (-oo,oo)# so every point at which #f'(x)# is undefined or #0# is a critical number.

Let's find #g'(x)# using the product rule:

#g'(x) = 2/5 x^(-3/5)(x-10) + x^(2/5)(1)#.

Let's now factor the denominators out of the expression:

#g'(x) = 1/5 x^(-3/5)[2(x-10)+5x]#.

We'll simplify the brackets and write the denominator as a denominator:

#g'(x) = (7x-20)/(5 x^(3/5)#.

#g'(x)# is not defined for #x=0# and is #0# at #x=20/7#.

Both are in #"Dom"(g)#, so both are critical numbers for #g#.