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

Scarlett Allison · Accepted Answer

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)#.

Aurora Alvarado · Answer

undefined To find the x-coordinates of all points of inflection, you need to find the second derivative of the function and solve for where it equals zero.

First derivative: $f'(x) = \frac{10}{3}x^{\frac{2}{3}} - \frac{20}{3}x^{-\frac{1}{3}}$

Second derivative: $f''(x) = \frac{20}{9}x^{-\frac{1}{3}} + \frac{20}{9}x^{-\frac{4}{3}}$

Setting $f''(x) = 0$, we get: $x = 0, \frac{1}{4}$

Now, to find discontinuities, look for values of x that make the function undefined. Here, there are none.

For intervals of concavity, analyze the sign of the second derivative in the intervals defined by the points of inflection. You'll find that $f''(x) > 0$ on $(-\infty,0)$ and $(\frac{1}{4},\infty)$, indicating concave up intervals, and $f''(x) < 0$ on $(0,\frac{1}{4})$, indicating a concave down interval.