AP Calculus BC - Second Derivatives

Determine the intervals on which the function is concave up:

$f=4\sin(\frac{x}{2})$

on the interval $0\leq x\leq 4\pi$

$(2\pi, 4\pi)$

The function is never concave up

$(0, 2\pi)$

$(0, \infty)$

Explanation

To determine the intervals on which the function is concave up, we must find the intervals on which the second derivative of the function is positive.

First, we must find the second derivative of the function:

$f'=2\cos(\frac{x}{2})$

$f''=-\sin(\frac{x}{2})$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Next, we find the values on the given interval for which the second derivative is equal to zero:

$-\sin(\frac{x}{2})=0$

$x=0, 2\pi, 4\pi$

We now use these values as bounds for the intervals on which we check the sign of the second derivative:

$(0, 2\pi), (2\pi, 4\pi)$

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, the function is concave up on the second interval, $(2\pi, 4\pi)$ .

Determine the intervals on which the function is concave up:

$f=4\sin(\frac{x}{2})$

on the interval $0\leq x\leq 4\pi$

$(2\pi, 4\pi)$

The function is never concave up

$(0, 2\pi)$

$(0, \infty)$

Explanation

To determine the intervals on which the function is concave up, we must find the intervals on which the second derivative of the function is positive.

First, we must find the second derivative of the function:

$f'=2\cos(\frac{x}{2})$

$f''=-\sin(\frac{x}{2})$

The derivatives were found using the following rules:

Next, we find the values on the given interval for which the second derivative is equal to zero:

$-\sin(\frac{x}{2})=0$

$x=0, 2\pi, 4\pi$

We now use these values as bounds for the intervals on which we check the sign of the second derivative:

$(0, 2\pi), (2\pi, 4\pi)$

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

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

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

Explanation

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

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

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

Explanation

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

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

Explanation

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

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

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

Explanation

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

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

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

Explanation

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

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

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

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

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

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

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

Explanation

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

Second Derivatives

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