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 .

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 negative, and on the second interval the first derivative is positive. Thus, the function is decreasing on the first interval, .

To determine the values at which the function has a relative maximum, we must determine the intervals on which the function's 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:

Note that we used factoring by grouping to determine the critical values.

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 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 first derivative changes from positive to negative at , so there exists a local maximum.

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, on the third interval, the first derivative is positive, on the fourth interval, the first derivative is negative, and on the fifth interval, the first derivative is positive. So, the function is decreasing on the 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 square root of a negative number isn't real, so the only critical values come from the first term.

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 function is decreasing on the second interval, .

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that .