All Calculus 1 Resources
Example Questions
Example Question #2 : Other Points
Find the critical points of
The critical points are complex.
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 #5 : Other Points
Find the critical points of
.
There are no critical points.
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 #2 : How To Graph Functions Of Points
Find the critical point(s) of .
and
and
and
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 #2581 : Calculus
Find the critical points of
.
There are no critical points
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 #2 : Other Points
Find all the critical points of
.
There are no critical points.
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 #3 : How To Graph Functions Of Points
Find the critical points of the following function:
To find critical points the derivative of the function must be found.
Critical points occur where the derivative equals zero.
Example Question #3 : Other Points
Determine the point on the graph that is not changing if .
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:
Limit does not exist.
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:
Example Question #12 : How To Graph Functions Of Points
Find
.
Limit does not exist.
To evaluate this limit, all we need to do is factor the numerator and then cancel out the factor that is in common using the following formula:
With a simple algebra trick, we will be able to easily plug in for :
Example Question #1561 : Functions
Find all points on the graph of where the tangent line is horizontal.
The tangent line is never horizontal for this graph.
To solve this problem, we need the chain rule, the derivative of the trigonometric function cosine, and the power rule.
First let's apply the chain rule, which states:
In this problem, and .
To find , we need the power rule which states:
To find , we need the derivative of cosine which states:
Plugging these equations into the chain rule we obtain:
To find all points where the tangent line is horizontal, we must first take the derivative of the function and then set it equal to zero:
Setting this equal to zero, we obtain:
Therefore, either or
Recall that from the unit circle, cosine equals zero at and sine equals zero at .
So, at every multiple of , either or .
Therefore, because at each multiple of , either or