To find when a function is concave, you must first take the 2nd derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, find the 2nd derivative:

\(\displaystyle y=\frac{1}{3}x^3+2x^2-5x-6\)

\(\displaystyle y'=x^2+4x-5\)

\(\displaystyle y''=2x+4\)

Set equal to 0 and solve:

\(\displaystyle 2x+4=0\)

\(\displaystyle 2x=-4\)

\(\displaystyle x=-2\)

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of -3 and 0.

\(\displaystyle y''=2x+4\)

\(\displaystyle y''(-3)=2(-3)+4=-2\)

\(\displaystyle y''(0)=2(0)+4=4\)

Since the value that is negative is when x=-3, the interval is decreasing on the interval that includes 0. Therefore, our answer is:

\(\displaystyle (-\infty,-2)\)

To test concavity, we must first find the second derivative of f(x)

\(\displaystyle \small f(x)=x^3+4x^2-5x\)

\(\displaystyle \small \small f'(x)=3x^2+8x-5\)

\(\displaystyle \small \small \small f''(x)=6x+8\)

This function is concave down anywhere that f''(x)<0, so...

\(\displaystyle \small \small \small \small f''(x)=6(x+\frac{4}{3})\)

So,

\(\displaystyle \small f''(x)< 0\)     for all       \(\displaystyle \small \small x< -\frac{4}{3}\)

So on the interval -5,-4 f(x) is concave down because f''(x) is negative.

To test concavity, we need to perform the second derivative test. Bascially, assuming that our function is twice-differentiable, anywhere that h"(t) is positive will be concave up, and anywhere h"(t) is negative is concave down. Begin as follows:

\(\displaystyle h(t)=t^4-5t^2+6t-14\)

\(\displaystyle h'(t)=4t^3-10t+6\)

\(\displaystyle h''(t)=12t^2-10\)

Next, we need to evaluate h"(t) on the interval [5,7]

\(\displaystyle h''(5)=12\cdot 5^2-10=290\)

\(\displaystyle h''(7)=12\cdot 7^2-10=578\)

So, our h"(t) is positive on the interval, and therefore h(t) is concave up.

The intervals where a function is concave up or down is found by taking second derivative of the function. Use the power rule which states:

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\)

\(\displaystyle y'=3x^2 +4x\)

\(\displaystyle y''=6x+4\)

Now, set \(\displaystyle y''\) equal to \(\displaystyle 0\) to find the point(s) of infleciton.

In this case, \(\displaystyle x=-\frac{2}{3}\).

To find the concave up region, find where \(\displaystyle y''\) is positive. This will either be to the  left of \(\displaystyle x=-\frac{2}{3}\)  or to the right of \(\displaystyle x=\frac{2}{3}\). To find out which, plug in a test point in each of those regions.

If we plug in \(\displaystyle x=-1\) we get \(\displaystyle y'' = -2\), which is negative, so that cannot be concave up.

If we plug in \(\displaystyle x=1\), we get \(\displaystyle 10\), which is positive, so we know that the region \(\displaystyle \left(-\frac{2}{3},\infty\right)\) will be concave up.

To find which interval is concave down, find the second derivative of the function.

\(\displaystyle y'=cos(x)\)

\(\displaystyle y''=-sin(x)\)

Now, find which \(\displaystyle x\) values in the interval specified make \(\displaystyle y''=0\). In this case, \(\displaystyle x=0\) and \(\displaystyle \pi\).

Now to find which interval is concave down choose any value in each of the regions

\(\displaystyle (0,\pi)\), and \(\displaystyle (\pi,2\pi)\)

and plug in those values into \(\displaystyle y''\) to see which will give a negative answer, meaning concave down, or a positive answer, meaning concave up.

A test value of \(\displaystyle \frac{\pi }{2}\) gives us a \(\displaystyle y''\) of \(\displaystyle -1\). This value falls in the range\(\displaystyle (0,\pi )\), meaning that interval is concave down.

The derivative of \(\displaystyle y=\sin{x}\) is

\(\displaystyle \frac{dy}{dx}=\cos{x}\)

The derivative of this is

\(\displaystyle \frac{d^2y}{dx^2}=-\sin{x}\)

This is the second derivative.

A function is concave down if its second derivative is less than 0.

\(\displaystyle -\sin{x}< 0\) whenever \(\displaystyle \sin{x}>0\)

This is true when:

 \(\displaystyle 0< x< \pi\)

To find the concavity of a graph, the double derivative of the graph equation has to be taken. To take the derivative of this equation, we must use the power rule,  

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

We also must remember that the derivative of an constant is 0.

After taking the first derivative of the equation using the power rule, we obtain 

\(\displaystyle y^{'}= x^2-5x+6\). 

The double derivative of the equation we are given comes out to 

\(\displaystyle y^{''}=2x-5\).  

Setting the equation equal to zero, we find that \(\displaystyle x=2.5\).  This point is our inflection point, where the graph changes concavity.  In order to find what concavity it is changing from and to, you plug in numbers on either side of the inflection point. if the result is negative, the graph is concave down and if it is positive the graph is concave up.  Plugging in 2 and 3 into the second derivative equation, we find that the graph is concave up from \(\displaystyle (2.5,\infty)\) and concave down from \(\displaystyle (- \infty ,2.5)\).

To find the invervals where a function is concave down, you must find the intervals on which the second derivative of the function is negative. To find the intervals, first find the points at which the second derivative is equal to zero. The first derivative of the function is equal to 

\(\displaystyle y'=9x^2+2x\).

The second derivative of the function is equal to 

\(\displaystyle y{}''=18x+2\).

Both derivatives were found using the power rule 

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

Solving for x, \(\displaystyle x=-\frac{1}{9}\). The intervals, therefore, that we analyze are \(\displaystyle \left(-\infty , -\frac{1}{9}\right)\) and \(\displaystyle \left(-\frac{1}{9}, \infty \right)\).

On the first interval, the second derivative is negative, which means the function is concave down. On the second interval, the second derivative is positive, which means the function is concave up. (Plug in values on the intervals into the second derivative and see if they are positive or negative.)

Thus, the first interval \(\displaystyle \left(-\infty , -\frac{1}{9}\right)\) is the answer. 

Points of inflection occur where there second derivative of a function are equal to zero. Taking the first and second derivative of the function, we find:

\(\displaystyle f'(x)=1+\cos(x)\)

\(\displaystyle f''(x)=-\sin(x)\)

To find the points of inflection, we find the values of x that satisfy the condition

\(\displaystyle -\sin(x)=0\).

Which occurs at \(\displaystyle x=0, \pm \pi, \pm2\pi, \pm3\pi...\)

Within the defined interval [-5, 5], there are three values: \(\displaystyle x=0, \pm\pi\).  These points are represented on the figure below as red dots.