AP Calculus BC - Inflection Points

$\begin{align*}&\text{Calculate any and all points of inflection for: }\&f(x)=10x + 2^{(6x)} + 2x^{2} - 2x^{3}\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative. For the function we're given, our}\&\text{first derivative is :}\&4x + 6\cdot 2^{(6x)}ln(2) - 6x^{2} + 10\&\text{And our second is:}\&36\cdot 2^{(6x)}ln(2)^{2} - 12x + 4\&\text{For a reminder of derivative rules:}\&d[b^{ax}]=ab^{ax}ln(b)dx\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x with a calculator yields:}\&x=-0.160-0.553i\&\text{Since we find no real roots, this function has no points of inflection.}\&\text{In this case, the function is always concave up.}\&\end{align*}$

$\begin{align*}&\text{Calculate any and all points of inflection for: }\&f(x)=x^{2} - 6x + 2x^{3} - 2x^{4} + 6\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&2x + 6x^{2} - 8x^{3} - 6\&\text{And our second is:}\&12x - 24x^{2} + 2\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=-0.132,0.632\&\text{We find real unique roots at the points }x=-0.132,0.632\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

$\begin{align*}&\text{Find any points of inflection for the function:}\&3^{(3x)} - 4x + 10x^{3} + 10x^{4} + 2x^{5}\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&3\cdot 3^{(3x)}ln(3) + 30x^{2} + 40x^{3} + 10x^{4} - 4\&\text{And our second is:}\&60x + 9\cdot 3^{(3x)}ln(3)^{2} + 120x^{2} + 40x^{3}\&\text{For a reminder of derivative rules:}\&d[b^{ax}]=ab^{ax}ln(b)dx\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=-0.154\&\text{We find one real unique root at the point }x=-0.154\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

$\begin{align*}&\text{Find any points of inflection for the function:}\&2^{(3x)} - 6x + 2x^{2} - 7x^{3} - 9x^{5} + 1\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative. For the function we're given, our}\&\text{first derivative is :}\&4x + 3\cdot 2^{(3x)}ln(2) - 21x^{2} - 45x^{4} - 6\&\text{And our second is:}\&9\cdot 2^{(3x)}ln(2)^{2} - 42x - 180x^{3} + 4\&\text{For a reminder of derivative rules:}\&d[b^{ax}]=ab^{ax}ln(b)dx\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x with a calculator yields:}\&x=0.214\&\text{We find one real unique root at the point }x=0.214\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

$\begin{align*}&\text{Find any points of inflection for the function:}\&7x - 2x^{2} - 2x^{3} + 6\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&7 - 6x^{2} - 4x\&\text{And our second is:}\&- 12x - 4\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=-0.333\&\text{We find one real unique root at the point }x=-0.333\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

Find the points of inflection on the function's domain:

$f=x^3+\frac{x^2}{4}$

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

There are no points of inflection

None of the other answers

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

Explanation

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:

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

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

The derivatives were found using the following rules:

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

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

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

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

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

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

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 $x=-\frac{1}{12}$ , because the second derivative did change sign at this point.

$\begin{align*}&\text{Calculate any and all points of inflection for: }\&f(x)=ln(5x) + 8x^{2} - 10x^{3} + 5x^{4} + 3\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative. For the}\&\text{function we're given, our first derivative is :}\&16x +\frac{ 1}{x}- 30x^{2} + 20x^{3}\&\text{And our second is:}\&60x^{2} -\frac{ 1}{x^{2}}- 60x + 16\&\text{For a reminder of derivative rules:}\&d[ln(ax)]=\frac{dx}{x}\&d[x^a]=ax^{a-1}dx\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x with a calculator yields:}\&x=-0.185,0.651,0.267-0.259i,0.267+0.259i\&\text{We find real unique roots at the points }x=-0.185,0.651\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

$\begin{align*}&\text{Determine what, if any exist, the points of inflection are for the function}\&y=8x - 4x^{2} + 3x^{3} + 4x^{4} - 3\end{align*}$

$\text{There are no points of inflection.}$

Explanation

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&9x^{2} - 8x + 16x^{3} + 8\&\text{And our second is:}\&18x + 48x^{2} - 8\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=-0.637,0.262\&\text{We find real unique roots at the points }x=-0.637,0.262\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

Which of the following is a point of inflection of on the interval $\left[\frac{\pi}{2},\pi\right]$ ?

$(\pi,12\pi)$

$(12\pi,\pi)$

$\left(\frac{\pi}{2},6\pi\right)$

$\left(6\pi,\frac{\pi}{2}\right)$

Explanation

Which of the following is a point of inflection of f(x) on the interval $\left[\frac{\pi}{2},\pi\right]$ ?

To find points of inflection, we need to know where the second derivative of the function is equal to zero. So, find the second derivative:

So, where on the given interval does ?

Well, we know from our unit circle that $sin(\pi)=0$ ,

So we would have a point of inflection at $x=\pi$ , but we still need to find the y-coordinate of our POI. find this by finding $f(\pi)$

$f(\pi)=5sin(\pi)+12(\pi)=12\pi$

So our POI is:

$(\pi,12\pi)$

Find all the points of infection of

$\ t_1=0 \ \ t_2=\frac{3\sqrt{10}}{10} \ \ t_3=-\frac{3\sqrt{10}}{10}$

$\ \ t_1=\frac{3\sqrt{10}}{10} \ \ t_2=-\frac{3\sqrt{10}}{10}$

$\ t_1=0$

There are no points of inflection.

Explanation

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

Now lets factor .

Now to find the points of inflection, we need to set .

From this equation, we already know one of the point of inflection, .

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

Recall that the quadratic equation is

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

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

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

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

Thus the other 2 points of infection are

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

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

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 $x=-1, x=-\frac{1}{2}, x=\frac{1}{2}, x=1$

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

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

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

Inflection Points

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