Differential Equations - CLEP Calculus

When the derivative of a function is equal to zero, that means that the point is either a local maximum, local miniumum, or undefined. The derivative of is . The derivative of the given function is

We must now set it equal to zero and factor.

Now we must plug in points to the left and right of the critical points to determine which is the local maximum.

This means the local maximum is at because the function is increasing at numbers less than -2 and decreasing at number between -2 and 6

The points where the derivative of a function are equal to 0 are called critical points. Critical points are either local maxs, local mins, or do not exist. The derivative of is . The derivative of the function is

Now we must set it equal to 0 and factor to solve.

We must now plug in points to the left and right of the critical points into the derivative function to figure out which is the local max.

This means that the function is increasing until it hits x=-6, then it decreases until x=1, then it begins increasing again.

This means that x=-6 is the local max.

At local maximums and minumims, the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative of the function.

The derivative of is . Thus the derivative of the function is

To find maximums and minumums we set it equal to 0.

So the critical points are at x=1 and x=2. To figure out the maximum we must plug each into the original function.

So the local max is at x=1.

To find the local max, you must find the first derivative, which is .

Then. you need to set that equal to zero, so that you can find the critical points. The critical points are telling you where the slope is zero, and also clues you in to where the function is changing direction. When you set this derivative equal to zero and factor the function, you get , giving you two critical points at and .

Then, you set up a number line and test the regions in between those points. To the left of -1, pick a test value and plug it into the derivative. I chose -2 and got a negative value (you don't need the specific number, but rather, if it's negative or positive). In between -1 and 1, I chose 0 and got a possitive value. To the right of 1, I chose 2 and got a negative value. Then, I examine my number line to see where my function was going from positive to negative because that is what yields a maximum (think about a function going upwards and then changing direction downwards). That is happening at x=1.

When the derivative of a function is equal to zero, that means that the point is either a local maximum, local miniumum, or undefined. The derivative of is . The derivative of the given function is

We must now set it equal to zero and factor.

Now we must plug in points to the left and right of the critical points to determine which is the local maximum.

This means the local maximum is at because the function is increasing at numbers less than -2 and decreasing at number between -2 and 6

The points where the derivative of a function are equal to 0 are called critical points. Critical points are either local maxs, local mins, or do not exist. The derivative of is . The derivative of the function is

Now we must set it equal to 0 and factor to solve.

We must now plug in points to the left and right of the critical points into the derivative function to figure out which is the local max.

This means that the function is increasing until it hits x=-6, then it decreases until x=1, then it begins increasing again.

This means that x=-6 is the local max.

At local maximums and minumims, the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative of the function.

The derivative of is . Thus the derivative of the function is

To find maximums and minumums we set it equal to 0.

So the critical points are at x=1 and x=2. To figure out the maximum we must plug each into the original function.

So the local max is at x=1.

By graphing the equation , we can see that there minimum at , and that the graph continues to rise in both directions around this point, so this must be a local minimum. We also know that the graph rises infinitely in both directions, so this must be the only local minimum.

Another way to identify the local minima is by taking the derivative of the function and setting it equal to zero.

Using the power rule,

we find the derivative to be,

.

From here we set the derivative equal to zero and solve for x. By doing this we will identify the critical values of the function

Now we will plug in the x value and find the corresponding y value in the original equation. We will also plug in an x value that is lower than the critical x value and a x value that is higher than the critical value to confirm whether we have a local minima or maxima.

Since both of the x values have a larger y value than the y value that corresponds to , we know that the minimum occurs at .

The points where the derivative of the function is equal to 0 are called critical points. They are either local maximums, local minimums, or do not exist. The derivative of is . The derivative of the function is

.

We must now set it equal to zero and factor to solve.

Now we must plug in points to the left and right of the critical points into the derivative function to find the local min.

This means the function is increasing until it hits x=2, then it decreases until it hits x=4 and begins increasing again. This makes x=4 the local minumum

The local maximums and minumums of a function are where the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative. Then we must set ot equal to 0 and solve. The derivative of is .

The critical points are at the above two points. To find the minimum we must plug both back into the origianl function.

Thus the local min is at x=-2.

To find the minimum of a function, start by finding the critical points of that function, or points where the derivative is equal to zero. Use the power rule to find the derivative:

Applying the power rule to the given equation, noting the constants in the first and second terms:

Then check to see if the critical point is a maximum, minimum, or an inflection point by taking the second derivative, using the power rule once again.

Because the second derivative is positive, the critical point is a minimum.

To find the point where the minimum occurs, plug back into the original equation and solve for .

Therefore, the minimum is

The local minimum of a function can be found by finding the derivative and graphing it. The point in which the x axis is crossed from below gives the x position where the local minimum is found. Taking the derivative:

The graph of the derivative is shown below:

As shown by the graph, the local minimum is found at x = -4.

By graphing the equation , we can see that there minimum at , and that the graph continues to rise in both directions around this point, so this must be a local minimum. We also know that the graph rises infinitely in both directions, so this must be the only local minimum.

Another way to identify the local minima is by taking the derivative of the function and setting it equal to zero.

Using the power rule,

we find the derivative to be,

.

From here we set the derivative equal to zero and solve for x. By doing this we will identify the critical values of the function

Now we will plug in the x value and find the corresponding y value in the original equation. We will also plug in an x value that is lower than the critical x value and a x value that is higher than the critical value to confirm whether we have a local minima or maxima.

Since both of the x values have a larger y value than the y value that corresponds to , we know that the minimum occurs at .

The points where the derivative of the function is equal to 0 are called critical points. They are either local maximums, local minimums, or do not exist. The derivative of is . The derivative of the function is

.

We must now set it equal to zero and factor to solve.

Now we must plug in points to the left and right of the critical points into the derivative function to find the local min.

This means the function is increasing until it hits x=2, then it decreases until it hits x=4 and begins increasing again. This makes x=4 the local minumum

The local maximums and minumums of a function are where the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative. Then we must set ot equal to 0 and solve. The derivative of is .

The critical points are at the above two points. To find the minimum we must plug both back into the origianl function.

Thus the local min is at x=-2.

To find the minimum of a function, start by finding the critical points of that function, or points where the derivative is equal to zero. Use the power rule to find the derivative:

Applying the power rule to the given equation, noting the constants in the first and second terms:

Then check to see if the critical point is a maximum, minimum, or an inflection point by taking the second derivative, using the power rule once again.

Because the second derivative is positive, the critical point is a minimum.

To find the point where the minimum occurs, plug back into the original equation and solve for .

Therefore, the minimum is

The local minimum of a function can be found by finding the derivative and graphing it. The point in which the x axis is crossed from below gives the x position where the local minimum is found. Taking the derivative:

The graph of the derivative is shown below:

As shown by the graph, the local minimum is found at x = -4.

First we must rearrange this separable differential equation so that we can place alone on one side and on the other side with the terms involving and any constants. We then integrate each side with respect to the appropriate variable and solve the result for to find the general solution of the differential equation:

Remember when integrating, we increase the exponent by one and then divide the whole term by the value of the new exponent. Will we need to integrate each term that contains in this fashion.

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the given initial condition to determine the value of the constant, which gives us the particular solution: