# Calculus 1 : Meaning of Functions

## Example Questions

### Example Question #51 : How To Find The Meaning Of Functions

Find the average value of  on the interval

Explanation:

To find the average value of a function on a closed interval [a,b], one uses the function

.

In this case, it is

, and using the general power rule for integral

and for natural log we know,

.

Applying these rules to our function, simplfies to

evaluated from .

This further simplies to

.

### Example Question #52 : Meaning Of Functions

Find the absolute maximum and minimum of the function  along the interval .

Absolute maximum: ; Absolute minimum

Absolute maximum: ; Absolute minimum

Absolute maximum: ; Absolute minimum

Absolute maximum: ; Absolute minimum

Absolute maximum: ; Absolute minimum

Absolute maximum: ; Absolute minimum

Explanation:

Find the x-value where

.

Thus, one of the solutions is .

Next  can be simplified to

.

Now plug in

into .

.

Thus the absolute maximum is at the point  and the absolute minumum is at the point .

### Example Question #51 : Meaning Of Functions

At what point on the graph of  is the tangent line parallel to the line

Explanation:

First one needs to find

.

Then find the slope of the line  (write it in the slope-intercept to do so).

The slope of the line is .

Set  and solve for x.

.

Plug in  into  to find the y-component of the point.

.

Thus the point is .

### Example Question #1773 : Functions

Find the limit:

undefined

Explanation:

If we plug  directly into the function in our limit, we get an indeterminate form, 0/0, so this is an excellent candidate for using L'Hopital's rule:

In this specific case, we get

Remember, the derivative of  is

If we try again to plug in 3, we get a more acceptable result of 8/6, which reduces down to 4/3.

### Example Question #1774 : Functions

The position of a particle at a point in time is given by the function .

At what point in time does the velocity of the particle reach a minimum?

Explanation:

We're given the particle's position function , we we'll first want to find the particle's velocity function. This can be found by taking the derivative of the position with respect to time:

Use the power rule to find the derivative:

Now, to find where the velocity reaches a minimum, take the derivative one more time to find the acceleration function:

Find the time where the acceleration is zero, i.e. the time when the direction of the acceleration changes:

Note that the sign of  changes from negative to positive at this point, indicating a minimum rather than a maximum.

### Example Question #51 : Meaning Of Functions

A huge container is used to hold all the confiscated candy at a local high school. Since everyone there loves sweets, the school is constantly confiscating candy, and dropping it into the container. At the end of the day, the container is emptied and starts empty the next day. The amount of candy in the container throughout the day is given by the following expression: , where t is in hours.

Assuming that there is a 10 hour school day, when is most of the candy confiscated?

Unable to answer with information given

Morning

Night

Constant from beginning of the day until the middle, then drops sharply

Mid-day

Mid-day

Explanation:

Although the candy is being added throught the day, the rate in which the candy is being added is not constant throught the 10 hour school day. If the amount of candy in the container is given by the expression , then in order to find the rate of candy being added to the container, you must take the derivative of this expression. The derivate of a function with respect to time tells you how that expression is changing over time. If you know how quickly the contents of the container are changing, then you can know when the rate of confiscation is at its highest. Thus, the first step in this problem is to find an expression for the rate of change:

.

This expression represents a parabola, as shown by the t2 value. At this point, there are three methods that could be used to find where the rate is the highest:

a) Theinformation ocould be extracted from the expression for the derivative. The negative t2 value indicates an upside down parabola. The (t-5) indicates that the parabola was shifted 5 spaces to the right. This means that the maximum occurs at t = 5, or midday

b) The derivative could be graphed from t = 0 to t = 10 to find the time at which the maximum value is reached

According to the graph, this peak happens at t = 5, or midday.

c) The derivative of the derivative could be taken and set to zero in order to find the value where the derivative reaches its local maximum.

### Example Question #51 : How To Find The Meaning Of Functions

A toy car is thrown straight upward into the air.  The equation of the position of the object is:

How long will it take for the toy car to hit the ground?

Explanation:

To find the time it takes for the toy car to hit the ground, we set the position equal to zero and we get  and .  At  is when the toy car was launched and at  is when the toy car reaches the initial position.

### Example Question #2801 : Calculus

A toy car is thrown straight upward into the air.  The equation of the position of the object is:

What is the instantaneous velocity of the toy car at  seconds?

Explanation:

To find the velocity of the toy car, we take the derivative of the position equation. The velocity equation is

Then we insert  seconds to find the instantaneous velocity.  A negative instantaneous velocity mean that the toy is slowing down.

### Find dv/dt if:

Explanation:

Solving for dv/dt, requires use of the chain rule:

This is one of the answer choices.

### Example Question #60 : Meaning Of Functions

Find

.

Explanation:

If you plug  into the equation, you get , meaning you should use L'Hopital's Rule to find the limit.

L'Hopital's Rule states that if

and that if

exists,

then

In this case:

Start by finding the derivative of the numerator and evaluating it at :

Do the same for the denominator:

The limit is then equal to the derivative of the numerator evaluated at  divided by the derivative of the denominator evaluated at , or .