Question

What are the points of inflection, if any, of `f(x)=x^7/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3`?

Answer 1

The only inflection point is `(1,-43/105)`

Explanation:

`f(x)=x^7/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3`

`f'(x) = x^6/6-(9x^5)/5+6x^4-(16x^3)/3`

`f''(x) = x^5-9x^4+24x^3-16x^2`

In order to determine where the concavity does change, we must first determine where it might change.

`f''(x)` is never undefined, so we need to solve `f''(x)=-`

`f''(x) = x^2(x^3-9x^2+24x-16)`

Use inspection or the rational zeros test to see that `1` is a zero of `x^3-9x^2+24x-16`.

Then use division or trial and error to factor

`x^3-9x^2+24x-16 = (x-1)(x^2-8x+16)`

`= (x-1)(x-4)^2`

This gives us:

`f''(x) = x^2(x-1)(x-4)^2`. So,

`f''(x) = 0` at `0, 1, "and " 4`.

But `0` and `4` are zeros of even multiplicity, so the sign of `f''(x)` does not change at those zeros.

The sign of `f''(x)` changes only at `x=1`.

( Note: this could also be determined by using a sign chart for `f''`.)

The inflection pint is the point `(1,f(1))`, so we do the arithmetic to find `f(1) = -43/105`, and

The inflection point is `(1,-43/105)`