Question

How many points of inflection does the function `f(x) = (pi/3)^((x^3)-8)` have?

Answer 1

One.

To find the points of inflection of any function, we compute the function's derivative then find the points where the derivative equals zero. In this case, we use the chain rule: let `u=x^3-8`, then

`f'(x)=d/dx (pi/3)^(x^3-8)=(d/dx (pi/3)^u)(d/dx (x^3-8))`
`=>f'(x)=(pi/3)^(x^3-8) ln(pi/3)(3x^2)`

There is no need to simplify this function any further. We know that for `f'(x)` to equal zero, then at least one of three things must happen:

`(pi/3)^(x^3-8)=0`, or
`ln(pi/3)=0`, or
`3x^2=0`.

Clearly, `ln(pi/3)` is not zero. Nor is `(pi/3)^(x^3-8)`, as any function `f(x)=a^(p(x))` is never zero for any polynomial `p(x)`. This leaves `3x^2`, and we know that `3x^2=0` only when `x=0`. Hence the one and only point of inflection is `x=0`.

Answer 2

Two points of inflections:
`x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1)`

Explanation:

A point of inflection can be obtained when `f''(x) = 0`.

Using the chain rule: let `u=x^3-8`, then

`f(x)=(pi/3)^(x^3-8)`
`f'(x)=d/dx (pi/3)^(x^3-8)=d/{du} (pi/3)^u {du}/{dx}`
`=(pi/3)^(x^3-8) ln(pi/3)3x^2`

`f''(x)=3ln(pi/3)(pi/3)^(x^3-8) x(x ln(pi/3) + 2) = 0`

Since:
`(pi/3)^(x^3-8)>0`
only possible solution is when:
`x(x ln(pi/3) + 2)=0`.

Hence two points of inflections:
`x=0 , f(0) = (pi/3)^(-8) " and " x=-2/ln(pi/3), f(-2/ln(pi/3)) = (pi/3)^(-8(1/(ln(pi/3))^3+1)`

Note that when `x=0, f(0)` it is also a stationary point.