To determine the local minima of the function, we must determine the points at which the function's first derivative changes from negative to positive.

The first derivative of the function is

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is positive. There is a sign change from negative to positive of the first derivative at , thus a local minima exists here. 

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Thus, the function is increasing on the first and third intervals, .

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

(Note that the method of completing the square was shown for solving for the critical values. One could use the quadratic formula as well.)

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative. and on the third interval, the first derivative is positive. Thus, the interval on which the first derivative is negative is the interval where the function is decreasing, .

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Note that factoring by grouping was used to find the critical values.

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. The intervals on which the function is decreasing are the intervals on which the first derivative was negative, .

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. The function is therefore increasing on two intervals, , on which we just showed the first derivative was positive.

At a relative minimum  of the graph , it will hold that  and . 

First, find . Using the sum rule,

Differentiate the individual terms:

Set this equal to 0:

Either:

, in which case, ; this equation has no real solutions.

 has two real solutions,  and . 

Now take the second derivative, again using the sum rule:

Substitute  for :

Therefore,  has a relative maximum at .

Now. substitute  for :

Therefore,  has its only relative minimum at .

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Thus, we know that the function is decreasing on the second interval, . 

To determine the values at which the function has a local maximum, we must determine the values at which the sign of the first derivative changes from positive to negative.

The first derivative of the function is equal to

and was found using the following rules:

, , 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, and on the third interval, the first derivative is negative. The first derivative changes from positive to negative at , thus there is exists a local maximum.

This is definitely a theoretical question. It is hard to imagine an equation that would have these properties, though it could be a piecewise function. This function has three critical points, which means that there are four regions where the rate of change could be positive or negative. (Where ). Since only one of the regions has a rate of change that is negative, the other three have positive rates of change. So this must mean that two regions that are next to each other will both have positive rates of change, making it look similar to a cubic function. So the two regions with f'(x)>0 will not have a maximum in between them, but the transition from the positive region to the negative region will have a maximum. The transition from the negative region to the positive will have a minimum not a maximum.

So: 

1 minimum

3 regions with f'(x)>0 

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

and was found using the following rules:

, 

Next, we must find the critical values, at which the first derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Therefore, the intervals on which the function is increasing are .