Question

How do you find the inflection points of `y=3x+root(5)((x+2)^3)`?

Answer 1

`(0,-6)`

Explanation:

`y=3x+(x+2)^(3/5)`

Recall that the inflection point is a point at which the curve changes concavity. To calculate the inflection point, we must find `y''` (the second derivative), set it equal to `0`, solve for `x` or determine `x` values at which `y''` does not exist (but `y` does) , and ensure that the concavity of `y` actually changes at these `x` values by testing values of `x` around our calculated `x` values by plugging them into `y''`: if `y''<0` on an interval, `y` is concave down on that interval; if `y''>0` on an interval, `y` is concave up on that interval.

`y'=3+(3/5)(x+2)^(-2/5)`
`y''=(3/5)(-2/5)(x+2)^(-7/5)`
`y''=(-6/5)(x+2)^(-7/5)`
`y''=(-6/(5(x+2)^(7/5)))`

There are no values of `x` for which `y''=0`; however, `y''` does not exist if `5(x+2)^(7/5)=0` because we cannot divide by `0`. So, let's solve `5(x+2)^(7/5)=0`:

`5(x+2)^(7/5)=0`
`(x+2)^(7/5)=0`
`(x+2)^((7/5)(5/7))=0^(5/7)`
`x+2=0`
`x=-2`

`y''` doesn't exist at `x=-2`, but `y` does. We now have the following intervals:

`(-∞, -2), (-2, ∞)`

For `x=-3`, `y''=(-6/5)(1/(-1)^(7/5))=(-6/5)(-1)=6/5>0`

Therefore, on the interval `(-∞, -2)`, `y` is concave up.

For `x=0`, `y''=(-6/5)(1/(2^(7/5)))<0`

Therefore, on the interval `(-2, ∞)`, `y` is concave down.

We have confirmed that the concavity of `y` changes at `x=-2`; therefore, there is an inflection point at `x=-2`. To find the `y`-value of the inflection point, calculate `y` at `x=-2`:

`y=3(-2)+(-2+2)^(3/5)=-6`

So, we have an inflection point at `(0,-6)`.

Answer 2

Compute `(d^2y)/(dx^2)`, the second derivative of `y`. Set this equal to 0, and solve for `x.` (The given curve does not have a point of inflection.)

Explanation:

When `y=0`, the curve has a root. At such a point, the function is neither positive nor negative.

When `dy/dx=0`, the curve has a horizontal tangent. At such a point, the function is neither increasing nor decreasing.

When `(d^2y)/(dx^2)=0`, the curve has an inflection point. At such a point, the function is neither concave up nor concave down.

Given `y=3x+(x+2)^(3//5)`, we compute both `y'` and `y'':`

`y'=dy/dx = 3 + 3/5(x+2)^(–2//5)`

`y'' = (d^2y)/(dx^2)=3/5(–2/5)(x+2)^(–7//5)`
`color(white)(y'' = (d^2y)/(dx^2))=–6/25 1/(x+2)^(7//5)`

Setting `y''=0`:

`0=–6/25 1/(x+2)^(7//5)`
`0=1/(x+2)^(7//5)`

In order for the fraction on the RHS to "equal" zero, `x+2` would have to be infinitely large `("+"//"–")`. Thus, there is no real point of inflection for the given `y.`

graph{3x+(x+2)^(3/5) [-2, 2, -1, 1]}