Calculus 1 : Graphing Functions

Study concepts, example questions & explanations for Calculus 1

varsity tutors app store varsity tutors android store

Example Questions

Example Question #101 : Graphing Functions

Find the  value(s) of the critical point(s) of

.

Possible Answers:

There are no real answers.

Correct answer:

Explanation:

In order to find the critical points, we must find  and solve for

Set  

Use the quadratic equation to solve for .

Remember that the quadratic equation is as follows.

 

, where a,b and c refer to the coefficents in the

equation  .

In this case, , , and .

After plugging in those values, we get

.

So the critical points  values are:

Example Question #21 : Points

Find the  value(s) of the critical point(s) of

.

Possible Answers:

Correct answer:

Explanation:

In order to find the critical points, we must find  and solve for .

Set   

Use the quadratic equation to solve for .

Remember that the quadratic equation is as follows.

 , where a,b and c refer to the coefficents in the equation  .

 

In this case, , , and .

After plugging in those values, we get. 

 

 

So the critical points  values are,

Example Question #3 : Other Points

Find the critical points of

Possible Answers:

 

The critical points are complex.

Correct answer:

Explanation:

First we need to find .

Now we set 

 

Now we can use the quadratic equation in order to find the critical points.

Remember that the quadratic equation is 

,

where a,b,c refer to the coefficients in the equation

 

In this case, a=3, b=6, and c=1. 

 

 

Thus are critical points are

 

Example Question #4 : Other Points

Find the critical points of

.

Possible Answers:

There are no critical points.

Correct answer:

Explanation:

In order to find the critical points, we need to find  using the power rule .

Now we set , and solve for .

Thus  is a critical point.

Example Question #5 : Other Points

Find the critical point(s) of .

Possible Answers:

 and 

 and 

 

 and 

Correct answer:

Explanation:

To find the critical point(s) of a function , take its derivative , set it equal to , and solve for .

Given , use the power rule

 to find the derivative. Thus the derivative is, .

Since :

The critical point  is 

Example Question #6 : Other Points

Find the critical points of

.

Possible Answers:

There are no critical points

Correct answer:

Explanation:

In order to find the critical points, we must find  using the power rule .

.

Now we set .

Now we use the quadratic equation in order to solve for .

Remember that the quadratic equation is as follows.

,

where a,b,c correspond to the coefficients in the equation 

.

In this case, a=9, b=-40, c=4.

 

 

Then are critical points are:

 

Example Question #7 : Other Points

Find all the critical points of

.

Possible Answers:

There are no critical points.

Correct answer:

Explanation:

In order to find the critical points, we first need to find  using the power rule ..

Now we set .

Thus the critical points are at

, and

.

Example Question #8 : Other Points

Find the critical points of the following function:

Possible Answers:

Correct answer:

Explanation:

To find critical points the derivative of the function must be found. 

 

Critical points occur where the derivative equals zero. 

Example Question #9 : Other Points

Determine the point on the graph that is not changing if .

Possible Answers:

Correct answer:

Explanation:

To find the point where the graph of  is not changing, we must set the first derivative equal to zero and solve for .

To evaluate this derivate, we need the following formulae:

Now, setting the derivate equal to  to find where the graph is not changing:

Now, to find the corresponding  value, we plug this  value back into :

Therefore, the point where  is not changing is 

Example Question #11 : Other Points

Find the limit: 

Possible Answers:

Limit does not exist.

Correct answer:

Explanation:

To evaluate this limit, we must use L'Hopital's Rule:

If  , take the derivative of both  and  and then plug in  to obtain 

We will also need the power rule, the derivative of the trigonometric function sine, and the chain rule.

Since when we plug in  in the numerator and denominator, we obtain a result of  , we can use L'Hopitals rule.

To take the derivative of the numerator we need the chain rule, the derivative of the trigonometric function sine, and the power rule.

Applying the chain rule to the numerator with  and , we see that:

 and .

Now plugging these into the chain rule, we obtain:

Now, to find the derivative of the denominator, we need the power rule again:

Now that we have found the derivative of the numerator and denominator, we can apply L'Hopital's Rule:

Learning Tools by Varsity Tutors