At what value of x does the graph of y=(1/x²)-(1/x³) have a point of inflection.

Solution:

Given y=(1/x²)-(1/x³)

We know that point of inflection means the curve changes its concavity from positive to negative.

It can change from convex to concave or vice versa.

That point occurs when the double derivative of f(x) is 0

Let's find derivative of f(x) = y=(1/x²)-(1/x³)

f'(x)=-2/x³ - (-3)/x⁴

Differentiating again, we get

(-2)(-3)/x⁴ -(-3)(-4)/x⁵ =0

6/x⁴ = 12/x⁵

x=2

Therefore, at x=2, we have a point of inflection.

At what value of x does the graph of y=(1/x²)-(1/x³) have a point of inflection.

Summary:

At x=2, the graph of y=(1/x²)-(1/x³) has a point of inflection.