Lexi Holmes

2023-03-25

How to find the inflection points for the given function $f\left(x\right)={x}^{3}-3{x}^{2}+6x$?

unnlattpdui

Beginner2023-03-26Added 5 answers

$f\left(x\right)={x}^{3}-3{x}^{2}+6x$

$f\prime \left(x\right)=3{x}^{2}-6x+6$

$f\prime \prime \left(x\right)=6x-6$

When the second derivative equals zero, inflexion points can be found.

For inflexion points, $f\prime \prime \left(x\right)=0$

$6x-6=0$

$x=1$

Test $x=1$ for concavity

When $x=0$, $f\prime \prime \left(0\right)=-6$

When $x=1$, $f\prime \prime \left(1\right)=0$

When $x=2$, $f\prime \prime \left(2\right)=6$

Hence, since there is a change in concavity, there is a point of inflexion at (1,4)

$f\prime \left(x\right)=3{x}^{2}-6x+6$

$f\prime \prime \left(x\right)=6x-6$

When the second derivative equals zero, inflexion points can be found.

For inflexion points, $f\prime \prime \left(x\right)=0$

$6x-6=0$

$x=1$

Test $x=1$ for concavity

When $x=0$, $f\prime \prime \left(0\right)=-6$

When $x=1$, $f\prime \prime \left(1\right)=0$

When $x=2$, $f\prime \prime \left(2\right)=6$

Hence, since there is a change in concavity, there is a point of inflexion at (1,4)

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