Is f(x) increasing, decreasing, or flat at ?

Recall that to find if a function is increasing or decreasing, we can use its first derivative. If f'(x) is positive, f(x) is increasing. If f'(x) is negative, f(x) is decreasing.

So, given:

We get

Then:

Therefore, f(x) is decreasing at the point, because f'(x) is negative.

Tell whether g(t) is increasing or decreasing on the interval [4,7]

To find increasing and decreasing, find where the first derivative is positive and negative. If g'(t) is positive, then g(t) is increasing and vice-versa.

Then,plug in the endpoints of [4,7] and see what you get for a sign.

So, since g'(t) is positive on the interval, g(t) is increasing.

To find the interval(s) on which the function is increasing, we must find the intervals on which the first derivative of the function is positive. 

The first derivative of the function is:

and was found using the rule

Now we must find the critical value, at which the first derivative is equal to zero:

Now, we make the intervals on which we look at the sign of the first derivative:

On the first interval the first derivative is positive, while on the second it is negative. Thus, the first interval is our answer, because over this range of x values, the first derivative is positive and the function is increasing. 

If  is continuous everywhere and always increasing (i.e.  for all ), then it must be true that after  has attained its root, it can never do so again because it can't "return" to the -axis. NOTE: this is not automatically true of functions that aren't continuous. As for the other choices, the possibility of at least having one more root is automatically false and a simple counterexample to the notion thay  has to have an inflection point is a simple increasing lines. It has a constant positive derivative, but possesses no upward or downward concavity and has no inflection points.

Take the first derivative of :

   by quotient rule

 is increasing whenever  is positive, that is, whenever both the numerator and denominator are of the same sign. The function  is certainly positive for all values of  greater than  because  and since 

is positive for all positive , it is increasing on the interval, too. It will never be negative. For the same reason, the numerator is always positive. With the numerator and denominator always positive everywhere on the given interval, the derivative is always positive and the function is always increasing. So for any interval of nonzero length within ,  is increasing.

NOTE: Interestingly the opposite of the choice  >  is also true.  on the entire interval because at , we have

. So the numerator is larger to begin with, and since:

  for all  (or any  for that matter), the derivative of the numerator is greater, too. This means the numerator will always be larger, so this condition coincides with the condition of  being positive.

To find the intervals on which the function is increasing, we must find the intervals where the first derivative is positive. To do this, we must find the first derivative, and find its critical values (at which the first derivative is equal to zero):

The derivative was found using the following rule:

Now, write the intervals of the function for which c is the upper and lower bound:

Note that at the critical value, the derivative is neither positive nor negative.

Now, we analyze the sign of the derivative within each interval; on the first interval, the derivative is always negative, but on the second interval, the first derivative is always positive. In other words, for this set of values -  -  the function is increasing. 

To determine the intervals on which the function is increasing, we must determine the intervals on which the first derivative of the function is positive. To start, we must find the first derivative:

The derivative was found using the following rule:

The first derivative is a positive constant, therefore the function is increasing on the entire domain, .

To find where the function is increasing, you must first find the derivative of the function so you can test critical points. The derivative of the function is . Then, set that equal to  to find the critical points. When you set that equal to , you get . Then set up a number line so you can test values to determine when the function is increasing and decreasing. We know it's changing direction at our critical point, . So let's pick a point to the left of  and plug it in to the derivative. I'll pick : . Since the answer is negative, we know that the function is decreasing. Pick a point to the right of 2. I'll pick 3: . Since the answer is positive, the function is increasing. Therefore, the function is increasing from .

To find the intervals on which the function is increasing, we must find the intervals on which the first derivative is positive. 

To start, we must first find where the first derivative of the function is zero:

The first derivative was found using the following rule:

Now, set this function equal to zero to get the critical values (values at which the first derivative is equal to zero):

Next, we create the intervals using the critical value as the upper and lower bound of the limits, respectively:

On the first interval, the first derivative is negative, but on the second interval, the first derivative is positive, meaning that on this interval the function is increasing. (Simply plug in any point on the interval into the first derivative function and check the sign.)

The answer is 

To solve, you must first differentiate the function once and then find where the derivative is positive. To differentiate, use the power rule:

Thus,

Now you must find where this is greater than 0, and therefore increasing.

Therefore, our answer is when x is greater than 2. Thus, .