Local Maximum - AP Calculus AB

To find the time at which the bean bag reaches its maximum height, we need to find the time when the velocity of the bean bag is zero.

We can find this by taking the derivative of the height position function with respect to time:

Setting this equal to 0, we can then solve for t:

The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:

To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:

or see that the equation factors into:

Either way, the only nonnegative root is 2.

To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:

Since the second derivative is positive, we know that this represents a local minima.

Recall that a function has critcal points where its first derivative is equal to zero or undefined.

So, given , we need to find v'(t)

Where is this function equal to zero? t=0 for one. We can find the others, but we really just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this case a saddle point.

A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.

First step is to find the derivative.

Find the critical points (when is or undefined).

Next, find at which of these two values changes from positive to negative. Plug in a value in each of the regions into .

The regions to be tested are ,, and .

A value in the first region, such as , gives a positive number, and a value in the second range gives a negative number, meaning that must be the point where the max occurs.

To find what the coordinate of this point, plug in in to , not , to get .

First, we need to find the critical points of the function by taking .

This is the derivative of a polynomial, so you can operate term by term.

This gives us,

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Solving for by factoring, we get

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This gives us critical values of 0 and . Since we are operating on the interval , we make sure our endpoints are included and exclude critical values outside this interval. Now we know the maximum could either occur at or . As the function is decreasing, we know at the max occurs at and that that value is .

Since a local maximum occurs at , this tells us two pieces of information: the derivative of must be zero at and the point must lie on the graph of this function. Hence we must solve the following two equations:

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From the first equation we get or .

To find the derivative we apply the quotient rule

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Solving , we get . Plugging this into the expression for gives .

To find the local maximum, we must find where the derivative of the function is equal to 0.

Given that the derivative of the function yields using the power rule . We see the derivative is never zero.

However, we are given a closed interval, and so we must proceed to check the endpoints. By graphing the function, we can see that the endpoint is, in fact, a local maximum.

First rewrite :

Use the multiplication rule to take the derivative:

To find the local extrema, set this to 0...

...and solve for ...

*

* Since we divided by , we have to remember that is a valid solution

Therefore, we know that we have two potential local extrema: and .

By plugging these in, we get two potential local extrema: and . Therefore, we know that the slope is positive between and . This means that can't be a local maximum, leaving only as a potential answer.

Next, we can find the slope at . It is:

This is negative, meaning that we go from a positive slope to a negative slope at , making it a local maximum.

To find the local maximum, first find the first derivative of the function.

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Then find all values of x for which the derivative equals 0 or is undefined. The derivative equals 0 when x=0 and is never undefined because the denominator is always greater than 0. Then, by picking points less than and greater than 0, we see that the function is increasing less than 0 and increasing greater than 0.

Therefore, it is a local maximum.

To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, .

Given the function, and applying the power rule we find the following derivative.

Check the -value at each endpoint and when the first derivative is zero, namely

The largest value is .

To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,

.

Given the function, and applying the quotient rule we find the following derivative.

when and when , which indicates that has a local maximum at .

To find the critical points of the graph, you first must take the derivative of the equation of the graph and set it equal to zero. To take the derivative of this equation, we must use the power rule,

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We also must remember that the derivative of an constant is 0. The derivative of the equation for this graph comes out to . Solving for when , you find that . The tricky part now is to find out whether or not this point is a local maximum or a local minimum. In order to figure this out we will find whether or not the slope is increasing towards this point or decreasing. Remember that the derivative of a graph equation gives the slope of the graph at any given point.

Thus when we plug in into the slope equation, we find that the slope has a positive value. This means that the slope is increasing as the graph leaves , meaning that this point is a local minimum, We plug in into the slope equation and find that the slope is negative, confirming that is the local minimum. That means that there is no local maximum on this graph.

Even though the first derivative () is at , there is no max or min because the function is increasing on both sides (derivative is positive on both sides). However, the function is changing its concavity (from negative to positive) at . We can tell because the second derivative () is also at , and it's going from negative to positive. Hole and asymptote are irrelevant. When the second derivative changes signs around a specific point we call this an inflection point. An inflection point describes a point that changes the concavity of a function.

To determine the maximum height of the ball, the first derivative of the function must be found and set equal to zero.

Test values to the left and right to confirm if this point is a maximum.

Because the derivative goes from positive to negative, we know the original function goes from increasing to decreasing and thus this point is a maximum.

Finally, to find the height of the ball at this point, plug in into the original function.

To find maxima and minima, find the coordinates of the points where the derivative is undefined or equal to zero. The derivative of p(x) is

Next set the derivative equal to zero and solve for x:

Finally we need to test the critical points in the original equation to determine which is a maximum.

Since the value of the function is greatest at x = -3, that is the x-coordinate of the maximum.

In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of where there are critical points. Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of in order to see whether or not the slope increases/decreases as is approaches or leaves that value.

In order to take the derivative of equation, the power rule must be applied, .

Taking the derivative, we find that the equation becomes .

Setting the equation equal to zero and solving for , we find that the critical points of this graph are . In order to determine whether or not these points are maxima or minima, we plug in numbers larger/smaller as well as inbetween the two critical points to find the behavior of the slope.

For this problem I will choose the numbers -5,0 and 1. These numbers are choosen since is approximately -4 and is approximately 0.25. By choosing numbers that are larger and smaller then the critical points as well as a number inbetween the two critical points, we can see the behavior of the slope throughout the entire graph.

Plugging in those numbers into , we find that makes positive, makes negative, and makes positive. This means that the slope is positive approaching and negative leaving it. The slope is negative approaching and positive leaving it. For a critical point to be the local maxima, its slope must be positive approaching it and negative leaving it, therefore is the local maxima in this graph.

and we need to find maxima. So lets take the first derivative by the product rule where . In this case, we can set . Next, we can find the derivatives of each . Now we assemble by the product rule above to get:

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Now we must find critical points (where , endpoints, or undefined values).

The domain of the function is all real numbers and the endpoints are and . So now we just need to find where . This will be equal to when and where . There is no value of we can choose such that so lets focus on . This is true where . We know that , so maybe we can use this to find values for which . We know that in the second quadrant, sine is positive and cosine is negative. We can check and verify that . The next value will occur in the 4th quadrant, when sine and cosine flip signs. So we can characterize these values that satisfy the equation as where .

Now we need to determine which of these critical points are maxima. so, the function is increasing from and we can show this is not a maxima. Since we're going to be testing a lot of points, let's take the second derivative and use the second derivative test. If , we have a max (think concave up). So .

At , , as (is in the second quadrant, where only sine is positive). Therefore, is a local maximum. The next value, , is in the fourth quadrant where cosine is positive, so the second derivative should be positive, meaning it should be a local minimum. The next point, should be another maximum, but it lies outside the domain. So the last endpoint, , is the last local maximum on the interval.

To find the local extrema of a function you must find the zeros of the derivative. The derivative of is , so we can calculate the derivative of our polynomial and pull out a constant, getting:

We can use the quadratic formula or just guess some easy values, and figure out that the zeros of this polynomial are x=6 and x=2. 2 is the only one of these in the interval stated, so we need to check and see if it's a maximum or a minimum. We have to plug at least three values into our function to see this, so we'll choose x=1, 2, 3.

We can see that as x approaches 2, f(x) is increasing, and as it moves away from 2, f(x) decreases.

So must be our local maximum.

To find the local maximum of the function, we must find the point at which the first derivative changes from positive to negative. To do this, we first must find the first derivative:

We found the derivative using the following rule:

Now, we must find the critical point(s), the point(s) at which the first derivative is equal to zero:

Now, we make our intervals over which to analyze the sign of the first derivative:

Over the first interval, the firt derivative is positive, and over the second interval, the first derivative is positive. Because the first derivative doesn't change from positive to negative, there are no local maxima.

To find the maximum, you must first find the derivative of the function, which is We found this derivative using the power rule which states,

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Then, you need to set that equal to zero to get your critical points.

It will be helpful to factor it:

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Then, set each expression to 0 to get those points:

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Then, put these points on a number line to then test the sign of the slope in between each point.

To the left of , plug in a value (say, 0) into the derivative. You get a positive value. In between and 1, the value is negative. To the right of 1, the value is positive. Then, to find the max, look for where the signs change from positive to negative.

That is occurring at .