Inflection points can only occur when the second derivative is zero or undefined. Here we have

$\displaystyle f'(x)=4x^3-9x^2, f''(x)=12x^2-18x=6x(2x-3)$. 

Therefore possible inflection points occur at $\displaystyle x=0$ and $\displaystyle x=\frac{3}{2}$. However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. Here we have

$\displaystyle f''(-1)=30, f''(1)=-6, f''(2)=12$. 

Hence, both are inflection points

Possible inflection points occur when $\displaystyle f''(x)=0$ . This occurs at three values, $\displaystyle x=-2, 0, 1$. However, to be an inflection point the sign of $\displaystyle f''(x)$ must be different on either side of the critical value. Hence, only $\displaystyle x=-2, 1$ are critical points.

A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function $\displaystyle y = x^3.$ 

The first derivative using the power rule  

$\displaystyle y=x^n \rightarrow y'=nx^{n-1}$ is,

$\displaystyle y' = 3x^2$ and the seconds derivative is $\displaystyle y'' = 6x.$ 

We then find where this second derivative equals $\displaystyle 0$.  $\displaystyle 6x = 0$ when $\displaystyle x = 0$.

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function $\displaystyle y'' = 6x$ does indeed change sign at, and only at, $\displaystyle x = 0$, so this is our inflection point.

In order to find the points of inflection, we need to find $\displaystyle f''(x)$ using the power rule, $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(x)=-x^3+x^2-x+1$

$\displaystyle f'(x)=-3x^2+2x-1$

$\displaystyle f''(x)=-6x+2$

Now we set $\displaystyle f''(x)=0$, and solve for $\displaystyle x$.

$\displaystyle -6x+2=0$

$\displaystyle -6x=-2$

$\displaystyle x=\frac{1}{3}$

To verify this is a true inflection point we need to plug in a value that is less than it and a value that is greater than it into the second derivative. If there is a sign change around the  point than it is a true inflection point.

Let $\displaystyle x=0$

$\displaystyle f''(0)=-6(0)+2=2$

Now let $\displaystyle x=1$

$\displaystyle f''(1)=-6(1)+2=-4$

Since the sign changes from a positive to a negative around the point $\displaystyle x=\frac{1}{3}$, we can conclude it is an inflection point.

In order to find the points of inflection, we need to find $\displaystyle f''(t)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(t)=t^4-2t^2+5t+100$

$\displaystyle f'(t)=4t^3-4t+5$

$\displaystyle f''(t)=12t^2-4$

Now to find the points of inflection, we need to set $\displaystyle f''(t)=0$.

$\displaystyle 12t^2-4=0$.

Now we can use the quadratic equation.

Recall that the quadratic equation is

$\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$,

where a,b,c refer to the coefficients of the equation  $\displaystyle at^2+bt+c=0$.

In this case, a=12, b=0, c=-4.

$\displaystyle t=\frac{-0\pm\sqrt{0^2-4(12)(-4)}}{(2)(12)}$

$\displaystyle t=\frac{\pm\sqrt{192}}{24}=\frac{\pm8\sqrt{3}}{24}=\frac{\pm\sqrt{3}}{3}$

Thus the possible points of infection are

$\displaystyle t_1=\frac{\sqrt{3}}{3}$

$\displaystyle t_2=-\frac{\sqrt{3}}{3}$.

Now to check if or which are inflection points we need to plug in a value higher and lower than each point. If there is a sign change then the point is an inflection point.

To check $\displaystyle t_1$ lets plug in $\displaystyle x=0, x=1$.

$\displaystyle f''(0)=12(0)^2-4=-4$

$\displaystyle f''(1)=12(1)^2-4=8$

Therefore $\displaystyle t_1$ is an inflection point. 

Now lets check $\displaystyle t_2$ with $\displaystyle x=0, x=-1$.

$\displaystyle f''(0)=12(0)^2-4=-4$

$\displaystyle f''(-1)=12(-1)^2-4=8$

Therefore $\displaystyle t_2$ is also an inflection point. 

In order to find the points of inflection, we need to find $\displaystyle f''(t)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(t)=t^5-3t^3+2t+100$

$\displaystyle f'(t)=5t^4-9t^2+2$

$\displaystyle f''(t)=20t^3-18t$

Now lets factor $\displaystyle f''(t)$.

$\displaystyle f''(t)=t(20t^2-18)$

Now to find the points of inflection, we need to set $\displaystyle f''(t)=0$.

$\displaystyle t(20t^2-18)=0$.

From this equation, we already know one of the point of inflection, $\displaystyle t_1=0$.

To figure out the rest of the points of inflection we can use the quadratic equation.

Recall that the quadratic equation is

$\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where a,b,c refer to the coefficients of the equation

$\displaystyle at^2+bt+c=0$.

In this case, a=20, b=0, c=-18.

$\displaystyle t=\frac{-0\pm\sqrt{0^2-4(20)(-18)}}{(2)(20)}$

$\displaystyle t=\frac{\pm\sqrt{1440}}{40}=\frac{\pm12\sqrt{10}}{40}=\frac{\pm3\sqrt{10}}{10}$

Thus the other 2 points of infection are

$\displaystyle t_2=\frac{3\sqrt{10}}{10}$

$\displaystyle t_3=-\frac{3\sqrt{10}}{10}$

To verify that they are all inflection points we need to plug in values higher and lower than each value and see if the sign changes.

Lets plug in $\displaystyle x=-1, x=-\frac{1}{2}, x=\frac{1}{2}, x=1$ 

$\displaystyle f''(-1)=20(-1)^3-18(-1)=-2$

$\displaystyle f''\left(-\frac{1}{2}\right)=20\left(-\frac{1}{2} \right )^3-18\left(-\frac{1}{2} \right )=6.5$

$\displaystyle f''\left(\frac{1}{2}\right)=20\left(\frac{1}{2} \right )^3-18\left(\frac{1}{2} \right )=-6.5$

$\displaystyle f''(1)=20(1)^3-18(1)=2$

Since there is a sign change at each point, all are points of inflection.

In order to find the points of inflection, we need to find $\displaystyle f''(x)$

$\displaystyle f(x)=x^2-2x+1$

$\displaystyle f'(x)=2x-2$

$\displaystyle f''(x)=2$

Now we set $\displaystyle f''(x)=0$.

$\displaystyle 0=2$.

This last statement says that $\displaystyle f''(x)$ will never be $\displaystyle 0$. Thus there are no points of inflection.

The points of inflection of a given function are the values at which the second derivative of the function are equal to zero.

The first derivative of the function is 

$\displaystyle f{}'(x)=3x^2+4x+1$, and the derivative of this function (the second derivative of the original function), is 

$\displaystyle f''(x)=6x+4$.

Both derivatives were found using the power rule 

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$. 

Solving $\displaystyle 0=6x+4$, $\displaystyle x=-\frac{2}{3}$.

To verify that this point is a true inflection point we need to plug in a value that is less than the point and one that is greater than the point into the second derivative. If there is a sign change between the two numbers than the point in question is an inflection point.

Lets plug in $\displaystyle {}x=-1$, 

$\displaystyle 6(-1)+4=-2$.

Now plug in $\displaystyle {}x=0$

$\displaystyle 6(0)+4=4$.

Therefore, $\displaystyle x=-\frac{2}{3}$ is the only point of inflection of the function.

In order to find all the points of inflection, we first find $\displaystyle f''(x)$ using the power rule twice, $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(x)=x^4+3x^3-4x+1$

$\displaystyle f'(x)=4x^3+9x^2-4$

$\displaystyle f''(x)=12x^2+18x$

Now we set $\displaystyle f''(x)=0$.

$\displaystyle 12x^2+18x=0$.

Now we factor the left hand side.

$\displaystyle x(12x+18)=0$

From this, we see that there is one point of inflection at $\displaystyle x=0$.

For the point of inflection, lets solve for x for the equation inside the parentheses. 

$\displaystyle \\ 12x+18=0 \\ \\ 12x=-18 \\ \\ x=-\frac{18}{12}=-\frac{3}{2}$

 In order to find all the points of inflection, we first find $\displaystyle f''(x)$ using the power rule twice $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(x)=x^4-x^2+100$

$\displaystyle f'(x)=4x^3-2x$

$\displaystyle f''(x)=12x^2-2$

Now we set $\displaystyle f''(x)=0$.

$\displaystyle 12x^2-2=0$

$\displaystyle \\ 12x^2=2 \\ \\ x^2=\frac{1}{6} \\ \\ x=\pm\sqrt{\frac{1}{6}}=\pm\frac{1}{\sqrt{6}}$

Thus the points of inflection are  $\displaystyle x=\frac{1}{\sqrt{6}}$ and $\displaystyle x=-\frac{1}{\sqrt{6}}$