Question

Find all possible inflection points of f(x) = -x^4 + 24x^2. Are there any actual inflection points of f(x)?

Answer 1

Given the function:

`f(x)=-x^4+24x^2`

You need to remember that the Inflection Points of a function are where it changes its concavity.

By definition, a function can have more than one inflection point:

1. You need to find:

`f^(\prime)(x)`

Remember the following Derivatives Power Rule:

`(d)/(dx)(x^n)=nx^(n-1)`

You get:

`f^(\prime)(x)=-4(x^(4-1))+(2)(24)(x^(2-1))`
`f^(\prime)(x)=-4x^3+48x`

2. Find the second derivative by derivating the first derivative. This is:

`f^{^(\prime)^(\prime)}(x)=(-4)(3)x^(3-1)+48`
`f^{^(\prime\prime)}(x)=-12x^2+48`

3. Find the third derivative by derivating the second derivative:

`f^{^{\prime^(\prime)^(\prime)}}(x)=(-12)(2)x^`
`f^{^{\prime^(\prime)^(\prime)}}(x)=-24x`

4.Find the roots of the second derivative:

- Set up that:

`-12x^2+48=0`

- And solve for "x":

`\begin{gathered} -12x^2=-48 \\ \\ x=\sqrt{(-48)/(-12)} \\ \\ x_1=2 \\ x_2=-2 \end{gathered}`

5. Substitute each root into the third derivative and evaluate:

`f^{^{\prime^(\prime)^(\prime)}}(2)=-24(2)=-48`

It is less than zero, therefore there is an inflection point in that x-value.

`f^{^{\prime^(\prime\prime)}}(-2)=-24(-2)=48`

It is positive output value, therefore there is an inflection point in that x-value.

6. Substitute those x-values into the original function and evaluate:

`f(2)=-(2)^4+24(2)^2=80`
`f(-2)=-(-2)^4+24(-2)^2=80`

Therefore, the inflections points are:

`\begin{gathered} (-2,80) \\ (2,80) \end{gathered}`

Hence, the answer is:

There are two inflection points:

`\begin{gathered} (-2,80) \\ (2,80) \end{gathered}`