At what value of x does the graph of y = 1/x² + 1/x³ have a point of inflection?

To find the point of inflection for the function y = 1/x² + 1/x³, we need to identify where the second derivative of the function changes sign. A point of inflection occurs when the concavity of the graph changes.

First, we’ll differentiate the function to find the first derivative:

y' = d/dx(1/x²) + d/dx(1/x³) = -2/x³ - 3/x⁴

Next, we simplify the first derivative:

y' = -2/x³ - 3/x⁴ = -2x - 3 / x⁴

Now, we compute the second derivative:

y'' = d/dx(-2/x³ - 3/x⁴) = 6/x⁴ + 12/x⁵

We can simplify the second derivative further:

y'' = 6/x⁴ + 12/x⁵ = 6x + 12 / x⁵

To find the points of inflection, we set the second derivative equal to zero:

6/x⁴ + 12/x⁵ = 0

This equation implies:

6x + 12 = 0

Solving for x gives:

x = -2

Next, we check the sign change around x = -2 to confirm there’s a point of inflection. We can pick values slightly less than and greater than -2 and observe the sign of the second derivative.

Thus, the graph of y = 1/x² + 1/x³ has a point of inflection at x = -2.