How do you find points of inflection and determine the intervals of concavity given #y=2x^(1/3)+3#?

Question

Ezekiel Alexander · Accepted Answer

Investigate the sign of #y''# #y' = 2/3x^(-2/3)# #y'' = -4/9x^(-5/3) = (-4)/(9x^(5/3))# Note that the sign of #x^(5/3)# is the same as that of #x#, so #y''# is positive left of #x=0# and the graph is concave up (convex) and #y''# is negative right of #0# and the graph is concave down (concave). The concavity changes at #x = 0# whic is in the domain of the function, so there is a inflection point at #x = 0# which make #y = 3# #(0,3)# is the only inflection point.

Emma Aldridge · Answer

undefined To find the points of inflection and determine the intervals of concavity for the function $y = 2x^\frac{1}{3} + 3$, follow these steps:

1. Find the first and second derivatives of the function.
2. Set the second derivative equal to zero and solve for $x$ to find the possible points of inflection.
3. Use the first derivative test or second derivative test to determine the concavity of the function in the intervals between the critical points found in step 2.
4. Determine the intervals of concavity based on the results from step 3.

Let's proceed with these steps:

1. Find the first derivative:
$$y' = \frac{2}{3}x^{-\frac{2}{3}}$$

2. Find the second derivative:
$$y'' = -\frac{4}{9}x^{-\frac{5}{3}}$$

3. Set the second derivative equal to zero and solve for $x$:
$$-\frac{4}{9}x^{-\frac{5}{3}} = 0$$
This implies $x$ can't be zero, so there are no points of inflection.

4. Determine the intervals of concavity:
Since the second derivative is negative for all $x$, the function is concave down for all $x$ values. Therefore, there are no intervals of concavity.

In summary, for the function $y = 2x^\frac{1}{3} + 3$, there are no points of inflection, and the function is concave down for all $x$ values.