$\displaystyle \begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&\frac{1}{x}- 16x - 24x^{2} + 5x^{4} - 1\\&\text{And our second is:}\\&20x^{3} -\frac{ 1}{x^{2}}- 48x - 16\\&\text{For a reminder of derivative rules:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-1.334,-0.471,1.697,0.054-0.210i,0.054+0.210i\\&\text{We find real unique roots at the points }x=-1.334,-0.471,1.697\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

$\displaystyle \begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&6 -\frac{ ln(2)}{2^x}- 9x^{2} - 4x\\&\text{And our second is:}\\&ln(\frac{2)^{2}}{2^x}- 18x - 4\\&\text{For a reminder of derivative rules:}\\&d[b^{ax}]=ab^{ax}ln(b)dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.192\\&\text{We find one real unique root at the point }x=-0.192\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

$\displaystyle \begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&20x +\frac{ 1}{x}- 21x^{2} + 20x^{3} - 4\\&\text{And our second is:}\\&60x^{2} -\frac{ 1}{x^{2}}- 42x + 20\\&\text{For a reminder of derivative rules:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.183,0.278,0.303+0.485i,0.303-0.485i\\&\text{We find real unique roots at the points }x=-0.183,0.278\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

$\displaystyle \begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&6x - e^{(-x)} + 30x^{2} + 28x^{3} + 3\\&\text{And our second is:}\\&60x + e^{(-x)} + 84x^{2} + 6\\&\text{For a reminder of derivative rules:}\\&d[e^{ax}]=ae^{ax}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.152\\&\text{We find one real unique root at the point }x=-0.152\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

The points of inflection are the points at which a function's second derivative changes in sign.

To start, we must find the function's second derivative:

$\displaystyle f'=3x^2+\frac{x}{2}$

$\displaystyle f''=6x+\frac{1}{2}$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we find the values at which the second derivative is equal to zero:

$\displaystyle 6x+\frac{1}{2}=0$

$\displaystyle x=-\frac{1}{12}$

Using this value as a bound, we create intervals on which to evaluate the sign of the second derivative:

$\displaystyle (-\infty, -\frac{1}{12})$. $\displaystyle (-\frac{1}{12}, \infty)$

Note that at the bounds of the intervals, the second derivative is neither positive nor negative. 

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. Thus, a point of inflection exists at $\displaystyle x=-\frac{1}{12}$, because the second derivative did change sign at this point.

To find inflection points (if there are any), we compute the second derivative of $\displaystyle f(x)$ and attempt to set it to $\displaystyle 0$. In this case:

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

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

so

 $\displaystyle x=\frac{1}{3}$.

Note: To confirm this is an inflection point (where the concavity of the function actually changes sign ––not simply touches $\displaystyle 0$ without changing sign––) we can take the derivative yet again:

$\displaystyle f'''(x)=-6$,

so of course, at our point of interest where $\displaystyle f''(x)=0$,

 $\displaystyle f'''\left (\frac{1}{3} \right )=-6$.

Since $\displaystyle f'''(x)$ is nonzero and negative, we know that the second derivative changes sign from positive to negative, and so at $\displaystyle x=1/3$ the function changes from concave up, to concave down.