How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #f(x)=2x^(5/3)-5x^(4/3)#?

Question

1356 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-14T17:49:34

Please see below.

#f'(x) = 10/3x^(2/3)-20/3x^(1/3)#

#f''(x) = 20/9x^(-1/3)-20/9x^(-2/3))#

# = 20/9x^(-2/3)(x^(1/3)-1)#

# = 20/9 * ((root(3)x-1)/(root(3)x^2))#

Sign of #f''#

The denominator is always positive and the numerator (hence the second derivative) is negative for #x < 1# and positive for #x > 1#.

There is a point of inflection at #x=1#

#f# is defined and continuous on #(oo,oo)#.

From the sign of #f''# we see that #f# is concave down (concave) on #(-oo,1)# and

#f# is concave up (convex) on #(1,oo)#.